% ElectricityBill/contents.tex
%
\chapter{\mychaptername}
\label{\here}
\chaptermark{Linear Functions}

\tocnotetoo{
We use an electricity bill as a hook on which to hang an introduction
to functions in general and linear functions in particular, in algebra
and in spreadsheets. Then we apply what we've learned to study taxes ---
sales, income and Social Security. You'll also find here a 
general discussion of energy and power.
}

\begin{goals}

\begin{goal}{directproportion}Understand direct proportion as a
linear equation with intercept 0.
\end{goal}

\begin{goal}{linearfunctions}
Study situations governed by linear equations.
\end{goal}

\begin{goal}{slopeintercept}
Stress the meaning and units of the slope and intercept.
\end{goal}

\begin{goal}{functions}
View functions as tables, graphs and formulas.
\end{goal}

\begin{goal}{linearexcel}
Construct flexible spreadsheets to model linear equations.
\end{goal}

\begin{goal}{incometax}
Understand the piecewise linear income tax computations.
\end{goal}

\begin{goal}{energyandpower}
Sort out the confusing distinction between energy and power.
\end{goal}

\end{goals}

\qrsection[directproportion]{Rates}
\begin{teacher}
In a course devoted more to real life applications than to modeling,
you can move directly to the section on taxes. There's no explicit
need there for the slope-intercept description of a linear model.
\end{teacher}

We began Chapter~\ref{Units} with a discussion of the relationship
\begin{center}
  distance = rate $\times$ time,
\end{center}
or, with units,
\begin{equation}\label{eq:rtd}
\text{distance (miles) = rate (miles/hour)} \times \text{time (hours)}.
\end{equation}

There we worked with particular numbers; now we want to
look at that relationship a little more generally.
If the rate is fixed (say 60 miles/hour), then we can find the
distance traveled whenever we know the time. To say that with some
algebra, write  $D$ for distance and $T$ for time. Then
%
\begin{equation*}
D = 60 \times T,
\end{equation*}
or, more generally,
\begin{equation*}
D = r \times T,
\end{equation*}
where $r$ is the rate of travel, in miles/hour. 

That formula says that distance traveled is
\emindex{proportional} to travel time.
The rate in miles per hour is
the \emindex{proportionality constant}.
If you drive for twice as long you go twice as far. 
If you drive ten times as long you go ten times as far. 

In \sref*{mpg} we introduced the units  (gallons per hundred miles) as
a useful way to measure automobile fuel economy: the amount of gas you
use is proportional to the distance you drive.
Let $F$ be the amount of gasoline you use, in gallons, to drive $D$ hundred
miles. Then
%
\begin{equation*}
F = r \times D
\end{equation*}
%
where the proportionality constant $r$ is the fuel
use rate, with units gallons per hundred miles. If you drive 10
times as far you use 10 times as much gas.
You don't use any gas at all just sitting in the driveway
(unless you're idling to warm up the car).

The proportionality constant is always a rate: it 
appears with units. In these examples the units are miles per hour
and gallons per 100 miles. 

The unit pricing discussion in \sref*{unitprices} provides more
examples of proportionality.

Finally, sales tax is computed as a proportion. If you spend twice as
much you pay twice as much tax. The tax rate is the proportionality
constant. If it's 5\% then you pay (five dollars of tax) per
(hundred dollars of purchase).

\qrsection[tamworth]{Reading your electricity bill}

The more electricity you use at home, the more you pay. But the
relationship isn't quite proportional. You don't pay twice as much to
use twice as much. Figure~\ref{tamworthbill} shows a simple sample
electricity bill based on one we found on a British website.
\begin{csmr}
Based on sample electricity bill at
\texttt{www.squashedfrogs.co.uk/} (broken link)
\access{August 31, 2015}.
\end{csmr}

\figfile{sampleelectricitybill-crop.pdf}
\begin{figure}
\centering
  \includegraphics[width=0.8\textwidth]{\thefigurefilename}
\caption{Tamworth electricity bill}
\label{tamworthbill}
\end{figure}
\figfile{}

This bill explains itself. We'll study it before we look
at a real one. Since it comes from Great Britain the costs are expressed
in pounds and pence rather than dollars and cents, and it comes once a
quarter (every three months) rather than once a month, but you can ignore
that while you read it --- from the bottom up.

The last line is the total bill, computed as
\begin{center}
\framebox{
\texttt{Cost of electricity + fixed charge} 
} .
\end{center}
Checking the arithmetic:
\begin{equation*}
\mbox{\textsterling} 34.62 + \mbox{\textsterling} 9.49
= \mbox{\textsterling} 44.11 .
\end{equation*}
The third line from the bottom explains the \textsterling 34.62:
\begin{center}
\framebox{
\texttt{Number of units $\times$ cost per unit} 
} .
\end{center}
That's our old friend proportionality.
The previous line gives the proportionality constant:
the cost per unit as 7.35 pence per unit. Later on in
the document we're told that a unit is just a kilowatt-hour,
abbreviated ``kwh''.
(It's too bad the bill talks about ``units'' instead
of just ``kwh'' since for us ``units'' has a more general meaning.)

There are 100 pence in a pound, so the cost per unit is
%
\begin{equation*}
0.0735 \frac{\text{\textsterling}}{\mbox{kwh}}.
\end{equation*}
%
(The computation would have been much
more complicated before February 15, 1971 --- the day England converted
from pounds/shillings/pence to decimal currency. See
\url{news.bbc.co.uk/onthisday/hi/dates/stories/february/15/newsid_2543000/2543665.stm}.)

In the current quarter this customer used 471 kwh of electricity ---
the difference between the meter reading before and after the quarter.

Here's all the arithmetic, with units:
\begin{equation*}
\mbox{\textsterling} 44.11 = 
0.0735 \frac{\mbox{\textsterling}}
{\mbox{\cancel{kwh}}} 
\times 471 {\mbox{\ \cancel{kwh}}}
+
\mbox{\textsterling} 9.49.
\end{equation*}

That English bill is easy to read. Figure~\ref{nstarbill} shows a real
one that's a little more complex, from NStar, in Boston.

\figfile{NStarElectricBill.pdf}
\begin{figure}
\centering
\framebox{
  \includegraphics[width=3.5in]{\thefigurefilename}
}
\caption{NStar electricity bill (2007)}
\figsource{Scan of one of the authors' bills - need to replace! Change
  to Eversource. Look   for image on line, or scan at higher resolution.}
\figcomment{Permission granted.}
\label{nstarbill}
\end{figure}
\figfile{}

We can identify the same two components. The 
fixed charge is the \$6.43 labelled ``Customer Charge.''
It's the part of the \$145.26 total  that does not 
depend on the amount of electricity used --- in this case, 813
kwh. The six lines on the bill that do depend on that contribute
%
\begin{align*}
(0.04432 & + 0.01039 + 0.00468 \\
         & + 0.00050 + 0.00250 + 0.10838) \times 813   \\
&     =  0.17077 \times 813   \\
&     =  138.83601 
\end{align*}
to the total bill, which is 
\begin{equation*}
\$145.26 = 
0.17077 \frac{\$}
{\mbox{\cancel{kwh}}} 
\times 813 {\mbox{\ \cancel{kwh}}}
+
\$6.43.
\end{equation*}
\noindent
Note that NStar rounded \$138.83601 down to \$138.83 rather than up to
the nearest penny. We should be grateful for small favors.

\qrsection[linearfunction]
{Linear functions}\index{function}
\index{linear function}

So far \CommonSense{} has called for hardly any algebra.
Now a little bit will come in handy.

Suppose you buy your electricity from NStar as in the example above 
and want to study how your
bill changes when you use different amounts of electricity.
The monthly \$6.43 Customer Charge does not change.
The rest of your bill is proportional to
the amount of electricity. The proportionality constant is
0.17077 \$/kwh in the sample bill. We will assume that it does
not change, although in fact it does change slightly when the electric
company changes its rates.

If in a given month you use $E$ kwh of electricity your total bill $B$ 
can be computed with the formula
\begin{equation*}
B\$ = 
0.17077 \frac{\$}
{\mbox{\cancel{kwh}}} 
\times E {\mbox{\ \cancel{kwh}}}
+
\$6.43 .
\end{equation*}

That formula captures how the dollar amount of your electricity
bill depends on the amount of electricity you use, measured in kwh.
The first term is the part that's proportional to the amount of
electricity used. The second term (the amount \$6.43) is fixed. It
represents the electric company's fixed costs: things like generating
the bill and mailing it to you and  maintaining the power 
lines on the street in front of your house. Those are expenses they
must cover even if you're on vacation and have turned off all the
appliances.

You \myindex{probably} encountered a similar formula once
in an algebra class --- it may look more
familiar without the units
\begin{equation*}
B = 0.17077 \times E + 6.43 .
\end{equation*}

It may look even more familiar if we call the variables by the
traditional names $x$ (for the independent variable) and $y$ (for the
dependent) instead of $E$ and $B$:
\begin{equation*}
y = 0.17077 x + 6.43.
\end{equation*}
This is a \emindex{linear function}, which standard algebra texts write in
\emph{slope-intercept} form
\begin{equation*}
y = m x + b.
\end{equation*}
In this example the {\emindex{slope} $m$ is 
$0.17077 \frac{\$}{\mbox{kwh}}$ and the \emindex{intercept} $b$ is \$6.43. 
For the English bill the slope is
$0.0735 \frac{\mbox{\textsterling}}{\mbox{kwh}}$ and the 
intercept is \textsterling{}9.49.%
\begin{teacher}
We usually treat the slope and intercept as given data, since that's
how they appear in the world. Computing the slope as $\Delta y /
\Delta x$ belongs in an algebra class that's leading to calculus, but
not here. 
\end{teacher}

There are many everyday examples where a
linear equation describes how a total is computed by
adding a fixed amount to a varying part that's a proportion.
The fixed amount is the intercept.
The proportionality constant is the slope.

\begin{itemize}

\item 
The most familiar examples are the ones where the intercept is 0:
all the ones in \sref*{directproportion}.

\item When renting a truck, the amount you pay is 
\begin{center}
(rate in dollars/mile) $\times$ (miles driven) + (fixed charge) .
\end{center}

\item Your monthly cell phone bill might be
\begin{center}
(rate in dollars per minute) $\times$ (number of minutes) +
(fixed fee) .
\end{center}

(A real cell phone bill will \myindex{probably} be more complicated,
perhaps with separate charges for phone minutes, text messages and
data transfer, perhaps with some of each kind of use built in to the
fixed fee.) 

\item If you work as a salesperson and your commission is 15\%  of
total sales 
your total wages are
\begin{center}
0.15 $\times$ (total sales) + (your base salary) .
\end{center}
\end{itemize}

The pattern is
\begin{center}
total = (rate) $\times$ (amount of some quantity) + (fixed constant).
\end{center}

In each case the slope is the rate and the intercept is the fixed
constant.  The units of a slope are always those of a rate.
In the truck example, the slope is the rate in dollars per mile; in
the cell phone example, the units of the slope are dollars per minute; 
in the salesperson example, the slope is 0.15 dollars of commission per
dollar of total sales. 

Think of the intercept as an initial or starting
value; it's what happens when the input is zero.  It has proper units
too --- in each of these examples that unit is dollars.
If you rent a truck but don't drive it anywhere, you still pay the fixed
charge. If you make no cell phone calls you still pay the
fixed fee. If you don't sell anything in a month, your commission is
\$0 but you still earn your base salary.   

\qrsection[linearexcel]{Linear functions in a spreadsheet}

In \sref*{tamworth} we saw that the amount you pay for
electricity in a month is a linear function of the amount you use.
In this section we'll use Excel to calculate electricity bills and to
draw a picture of the results.

Figure~\ref{fig:TamworthElectricScreenShot1} is a screen shot of the 
Excel spreadsheet \link{TamworthElectric.xlsx}. 
We put the slope 0.735 in
cell \cell{C4}, with its units  {\textsterling}/kwh in cell \cell{D4}.  We put the intercept  9.49 in cell
\cell{C5} and the units  (\textsterling) in cell \cell{D5}.

\figfile{TamworthElectricScreenShot1-2019.pdf}
\begin{figure}
\centering
  \includegraphics[width=4in]{\thefigurefilename}
\caption{Tamworth electricity bill}
\figsource{Excel screen shot}
\label{fig:TamworthElectricScreenShot1}
\end{figure}
\figfile{}

Then we entered column labels in cells \cell{B7:C8} and
a few values in rows \cell{9} through 
\cell{13}  in column \cell{B}. Finally, we asked Excel to calculate
the electricity bills in column \cell{C}. To do that, we started with 
the formula
\displayexcel{
%	=C\$4*B9+C\$5
	=C4*B9+C5
}
\noindent in cell \cell{C9}. The \excel{=} sign tells Excel to
multiply the numbers in cells \cell{C4} and \cell{B9} and add the
number in \cell{C5}. The result is \excel{9.49}, as expected.

The next step is to copy that formula from \cell*{C9} to
\cell*{C10}. With any luck Excel will guess what we want to do, change
\excel{B9} to  
\excel{B10}, compute
\displayexcel{
%	=C\$4*B9+C\$5
	=C4*B10+C5
}
\noindent
and display \excel{44.11}. But that's not what happens! Excel
shows \excel{4469.79} instead! If you look at the contents of
\cell*{C10} you will find
\displayexcel{
	=C5*B10+C6
}
so Excel added 1 to the row numbers for cells \cell{C4} and \cell{C5}
as well as for \cell{B9}. 
There's nothing in \cell*{C6}. Excel treats
that as a zero and adds it to $471 \times 9.49$ to get $4469.79$.

That's not what we want. Changing \cell{B9}
to \cell{B10} is right, but we want Excel to leave the
references to cells \cell{C5} and \cell{C6} alone. The trick
that makes that happen is to put a \cell{\$} in front of the \cell{5}
and the \cell{6}. This is not something you could have figured
out. There's no particular reason why this trick should work. 
Just remember it. The right formula to use in
\cell*{C9} is 
%
\displayexcel*{
	=C{\textbf\$}4*B9+C\textbf{\$}5
}
\noindent
When we copy that formula from \cell{C9} to cells
\cell{C10:C13} we get Figure~\ref{fig:TamworthElectricScreenShot1}.

Figure~\ref{fig:TamworthElectricScreenShot2} is a screen shot
of the same spreadsheet --- after we asked Excel to show the
formulas for each cell instead of the values. 

\figfile{TamworthElectricScreenShot2-2019.pdf}
\begin{figure}
\centering
  \includegraphics[width=4in]{\thefigurefilename}
\caption{Tamworth electricity bill --- formulas}
\figsource{Excel screen shot}
\label{fig:TamworthElectricScreenShot2}
\end{figure}
\figfile{}

In Chapter~\ref{IncomeDistribution} we learned how to use Excel to
draw bar charts and histograms so that we could visualize data
organized into categories. The $x$-axis displayed category names, with
corresponding values on the $y$-axis. That won't work for the data in
Figure~\ref{fig:TamworthElectricScreenShot1}, since there both the
$x$- and $y$-axes have numerical values. Instead, after selecting
cells \cell{B7:C13} we must ask Excel for a chart of
type\index{chart type} \excel{XY(Scatter)}\index{scatter
  chart}. Figure~\ref{fig:linearGraph} shows the result. 

The graph is a straight line --- 
that's why the function is called ``linear''. The slope tells us how
steep the line is and the intercept tells us where it crosses the
vertical axis --- in this case at the value 
\textsterling{}9.49, the total bill when you use no electricity at
all.
\begin{teacher}
Since the entries in the Tamworth Electricity Bill table are out of
order, Excel has drawn some of the segments in the graph twice. You can
see that if you look carefully. You might want to point this out, or not.
\end{teacher}

\figfile{TamworthElectricBillExcel-2019.pdf}
\begin{figure}
\centering
  \includegraphics[width=0.8\textwidth]{\thefigurefilename}
\caption{Tamworth electricity bill --- chart}
\label{fig:linearGraph}
\end{figure}
\figfile{}

Excel will let you change the type of a chart once it's built. If you
change the chart in Figure~\ref{fig:linearGraph} to a 
\excel{Line Chart}\index{line chart} Excel 
will use the data in the column \cell{B} as category labels rather
than as the numbers of kilowatt-hours. 
It will space them evenly along the $x$-axis, whatever their
values, and draw the nonsense you see in Figure~\ref{nonsense}.

\figfile{TamworthElectricBillLineChart-2019.pdf}
\begin{figure}
\centering
  \includegraphics[width=0.8\textwidth]{\thefigurefilename}
\caption{How NOT to draw a line}
\label{nonsense}
\figsource{REDRAW hires. Chart from Excel spreadsheet we built.}
\end{figure}
\figfile{}

If you change the chart type to scatter you get two
scatters, one for each column. You can get 
Figure~\ref{fig:linearGraph} only if you start with a scatter plot.
If you select the two columns of data and build a line chart first,
things are even worse. 
Excel thinks each row is a category for which you have two
pieces of data. It labels the categories 1, 2, \ldots and shows a line 
for each.
Try this and see what happens.

\qrsection[comparelinear]{Which truck to rent?}

Table~\ref{table:TruckRental} shows the cost of renting a truck for
one day in Boston in March of 2015. Four companies offer equivalent
models. Which one should you choose?

\begin{center}
\begin{tabular}{cS[table-format=2.0]S[table-format=2.2]S[table-format=2.2]S[table-format=2.2]}
\toprule
           & {Watertown} & {U-Haul} & {Budget} & {Enterprise} \\
\midrule
fixed cost & 79 & 29.95 & 29.95 & 59.99 \\
\$/mile & 0 & 1.39 & 0.99 & 0.59 \\
\bottomrule
\end{tabular}
\captionof{table}{One day truck rental costs}
\label{table:TruckRental}
\end{center}

It's clear from the numbers that for a very few miles Budget will be
cheapest, since it's tied with U-Haul for the lowest fixed cost
and charges less per mile. For a really long move Watertown will be
best since there is no mileage charge. 
The Excel chart in
Figure~\ref{fig:TruckRental} tells the whole story. Budget is cheapest
up to about 50 miles. For longer trips, choose Watertown. It  never
makes financial sense to rent from U-Haul or Enterprise.

The figure is a good reminder of the meaning of slope and intercept
for straight line graphs. A line crosses the
vertical axis at the intercept. In this example intercepts represent the fixed
costs. The units of the intercept are dollars --- the units on the
$y$-axis. The slope of a line measures how steep it is.
That's particularly visible when you compare U-Haul
and Budget, which have the same intercept but different slopes ---
1.39 dollars/mile and 0.99 dollars/mile. The units of the slope are
always (units on $y$-axis)/(units on $x$-axis),
The line for Watertown is horizontal since its slope is 0 dollars/mile. 

\figfile{TruckRentalCropped.pdf}
\begin{figure}
\centering
  \includegraphics[height=3in]{\thefigurefilename}
\caption{Comparing truck rental costs}
\label{fig:TruckRental}
\end{figure}
\figfile{}

The spreadsheet with that figure is at
\link{TruckRental.xlsx}. Figure~\ref{fig:TruckRentalScreenshot}
shows how we arranged the formulas in the spreadsheet to compute the
total cost for each company.
Column \cell{A} lists the mileages we're considering, from
0 in \cell*{A11} to 100. The formula \excel{=A11+10} in
\cell*{A12} fills column \cell{A} when we copy it to cells
\cell{A13:A21}. The formula in \cell*{B11} is
\displayexcel*{
=B\$7+\$A11*B\$8
}
It uses three \excel{\$} signs to keep Excel from adjusting
references for rows \cell{7} and \cell{8} and for column
\cell{A}. That allowed us to copy it to all of the range
\cell{B11:E21}.

\figfile{TruckRentalScreenshot-2019.pdf}
\begin{figure}
\centering
  \includegraphics[height=2in]{\thefigurefilename}
\caption{Computing truck rental costs}
\label{fig:TruckRentalScreenshot}
\end{figure}

The problem we've just solved is typical of situations where you have
to decide among options, some with small startup cost but a
high ongoing rate, others the reverse. Here are some examples:

\begin{itemize}
\item Insulate your house (high initial investment compared to doing
nothing) in order to pay less for heat in the winter (lower rate for
use).

\item Buy a hybrid instead of a conventional car (higher initial cost,
lower rate of fuel consumption).

\item Buy energy efficient light bulbs (more expensive to start with,
but they use less electricity to run),

\item Select a phone plan with unlimited text messaging (more
expensive than pay-as-you-text, but the slope (\$ per text) is zero).
\end{itemize}

Each of these can be thought of as looking at linear equations to
see where their graphs cross. You can do that by building a table of
values  or by drawing the graphs (in Excel, or with pencil
and paper) or by writing down the equations and solving them with
remembered algebra, or by \myindex{guess-and-check}. 

But you can't rely on just this mathematics to make a decision. There
are always other important things to think about. How long does the
more expensive purchase last? Can you afford the initial high payment?
If so, what else might you rather do with that money? Do you need to
take depreciation or inflation into account?


\qrsection[energypower]{Energy and power}

How much electrical energy does a 100 watt light bulb use? That
depends on how long it's on. When it's switched off, it doesn't
use any at all. If it's on for two hours it must use twice as much as
it does in one hour. 
Figure~\ref{fig:TamworthPower} (from the second page of the Tamworth
bill) shows the 
proportion lurking there.

\figfile{TamworthPowercropped.pdf}
\begin{figure}
\centering
  \includegraphics[height=20mm]{\thefigurefilename}
\caption{How much electrical energy?}
\figsource{From the Tamworth .doc file.}
\label{fig:TamworthPower}
\end{figure}
\figfile{}

The second line in that figure displays the units for the
quantities in the first line. Time is measured in hours,
of course. Power is measured in kilowatts. Energy is
measured in kilowatt-hours: the product of the units for power
and for time.

That tells us right away that energy and power are not the
same thing. Comparing the figure to Equation~\ref{eq:rtd}, you can see
that energy is like distance --- a thing that's consumed or
traveled, while power is like speed --- the rate at
which energy is used or distance covered.%
\begin{teacher}
We're undecided about how much time to spend on power and energy.
It takes a lot of teaching time (and energy) to
convey the distinction convincingly in class. Perhaps those resources
are better spent on other parts of the curriculum. But the topic is
important and ultimately interesting, because students care about
climate change and alternative power sources. We have included
problems that explore the issue further.

If you do choose to spend more time on the difference between power
and energy, and the confusion because the name ``watt-hour'' contains
a unit of time, consider discussing the light-year, which is a measure
of distance, not time. So ``light-years ago'' is never right. The
knot --- one nautical mile per hour --- is also a rate, like power, that
doesn't mention time. 
\end{teacher}

If you turn a 100 watt light bulb on for 2 hours the
formula in Figure~\ref{fig:TamworthPower} tells you how much
electrical energy you use:
\begin{align}
100 \text{ watts} \times 2 \text{ hours } 
& = 200 \text{ watt-hours} \label{eq:energypower1}\\
& = 0.2 \text{ kilowatt-hours} \label{eq:energypower2}
\end{align}
(\ref{eq:energypower1}) is just multiplication.
(\ref{eq:energypower2}) changes watt-hours to kilowatt-hours.

How much does it cost to leave a 40 watt bulb on all the time in
your basement for a year? There are about 9,000 hours in a year. That
is 9 kilo-hours, so you'll use about
%
\begin{equation*}
40 \text{ watts} \times 9 \text{ kilo-hours} = 360 \text{ kilowatt-hours}.
\end{equation*}
If you pay \$0.10 per kwh for electricity it will cost you about \$36 per
year to guarantee that you don't fall down the basement steps in the dark.

The electrical energy that flows through the wires in your house to
your appliances \myindex{probably} comes to you from a power plant, which
might be burning coal or natural gas or extracting energy from nuclear
fuel. (You might have a wind turbine in your neighborhood, or a hot
spring, or solar panels on your roof, but these are unlikely power
sources for most people.)\index{solar energy}
So power plants produce energy, not power. The power of a power
plant is the rate at which it can produce energy, so
 ``energy plant'' would be a better name than 
``power plant''.

The website for Chicago's Cook Nuclear Plant says that
\begin{quotation}
The 1,048 net megawatt (MW) Unit 1 and 1,107 net MW Unit 2
  combined produce enough electricity for more than one and one half
  million average homes.%
\begin{csmr}
The Cook Nuclear Plant.
\url{www.cookinfo.com/cookplant.htm}
\access{July 17, 2015}.
\csmrcomment{29 words fair use}
\end{csmr}
\end{quotation}

Let's check this. The combined total power is 2,155 megawatts.  This
is the rate at which that plant produces electrical energy when
it is running at full power. (When it's not running it's still just as
powerful, but not producing any energy.) When it's running, how many
average homes could it produce electricity for?

If the Cook plant ran all year (about 9,000 hours) it would produce
\begin{align*}
2,155 \text{ megawatts} \times 9,000 \text{ hours}
& \approx 18,000,000 \text{ megawatt-hours} \\
& = 18,000,000,000 \text{ kilowatt-hours}
\end{align*}
of electrical energy.
Googling ``average household electricity usage'' finds
\begin{quotation}
 6,000 kwh per household per year for 3 residents average per
household
\end{quotation}
from
\url{www.physics.uci.edu/~silverma/actions/HouseholdEnergy.html}.
The source is a physics professor's website, so it's \myindex{probably}
reliable. 
At 6,000 kwh per household per year Cook could power 3 million
homes. The quotation claims half that, so it's clearly in the right
ballpark. The 6,000 kwh per household per year is a southern
California average --- households in northern Illinois might well use
more electricity.

Energy comes in many forms besides electric. The Cook plant
converts the energy in its nuclear fuel to electricity. 
Driving a car uses the energy stored in the gasoline.
Running a \myindex{marathon} uses the energy in the food you eat.
Each form of energy has its own units. We've seen that electrical
energy is measured in kilowatt-hours.
If you cook on a gas stove, the energy in the gas is measured in
\emph{therms}.\index{therm} The energy in the oil that heats  
your house is measured in \emindex{British Thermal Units}
\index{Btu} or Btus. The energy in the
food you eat is measured in \emph{calories}\index{calorie}. Physicists
measure energy in \emph{ergs} or \emph{joules}; you rarely see those
units in everyday life. You can look up conversion factors for these
units --- for example, the energy in a barrel (42 gallons) of oil is
about 5.8 million Btu, which is
equivalent to 1700 kilowatt-hours. So it would take about a fifth of a
barrel to keep that 40 watt light bulb burning for a year.

Converting among the units for energy is just like converting among
the units for length (meters, feet, yard, miles, \ldots). You can use
a table, an online calculator like the one at the National Institute
of Standards and Technology
(\url{physics.nist.gov/cuu/Constants/energy.html}) or the
Google calculator.

Possibly the most interesting energy conversion is the one that
Einstein discovered in 1905: mass and energy are the same thing,
measured in different units. The conversion factor is the square of
the speed of light --- hence the famous equation
\begin{equation*}
e = mc^2.
\end{equation*}

To see that at work, look again at the yearly energy output of the
Cook plant. The Google calculator tells us that
\gc*{
18 000 000 000 kilowatt-hours = $6.48 \times 10^{16}$ joules
}
The National Institute of Standards and Technology website says
that corresponds to a mass of about 0.72 kg, which is 720
grams. That means just about 1.6 pounds of matter must be converted to
energy to power millions of Chicago homes for a year.
The Google calculator does not know Einstein's equation, so
it wouldn't convert kwh to grams directly!

%\input{\here/tax2019}
\qrsection[taxes]{Federal payroll taxes}

\maasection{Income tax}

Taxes are a part of life (the only other certainty is death),
so it's only common sense to learn how they work. In 
\sref*{directproportion} we studied sales taxes. 
Cities and states collect them; they are computed
as a percentage of the purchase price. In this section we'll explain
two important federal taxes that depend on your income, not on how you
spend it. 

Federal income tax is not simply a proportion of your income.
It's a \emph{progressive graduated tax}\index{progressive
  tax}\index{graduated tax}. When you make more money you not only pay
more tax, some of your income may be taxed at a higher rate. 
Table~\ref{table:2019taxtable} shows the 2019 \emph{tax brackets}
for single taxpayers. 

\begin{table}
\centering
\begin{tabular}{r@{\ --\ }lS[table-format=2.1]}
\toprule
\multicolumn{2}{c}{Bracket (\$)} & {Marginal Tax Rate (\%)} \\ 
\midrule
0  &  9,700 & 10 \\
9,701  &  39,475 & 12 \\
39,476  &  84,200  & 22 \\
84,201  &  160,725 & 24 \\
160,726  &  204,100  & 32 \\
204,101 & 510,300 & 35  \\
510,301 & & 37 \\
\bottomrule
\end{tabular}
\caption{2019 single taxpayer brackets and rates}
\tablesource{Public information.}
\label{table:2019taxtable}
\end{table}

That tells you that the first \$9,700 of
your income is taxed at 
10\%. If you make exactly that much, you pay \$970 in
tax. If you make more, the extra income is taxed at a higher rate ---
you have moved to a higher \emph{tax bracket}\index{tax bracket}.
For example, if you make between \$9,700 and \$39,475 you will pay
\$970 for the first \$9,700 and
12\% of
the amount you earn over \$9,700. If you earn more than
\$39,475 you start paying at a 22\% rate on the extra.

Let's try an example.  If your taxable income is \$50,000, then your
total tax is 
\begin{align}\label{eq:fiftyK}
\text{total tax} & =  0.10 \times \$9,700
+ 0.12 \times (\$39,475 - \$9,700) \\
& \quad
+ 0.22 \times (\$50,000-\$39,475) \notag \\
& =  \$970.00 + \$3573.00 + \$2315.50 \notag \\
& =  \$6858.50 \ . \notag
\end{align}

Note carefully that when you are in a higher tax bracket the higher
rate applies only to the extra income. The taxpayer in this
example is in the 22\% bracket, but that rate applies only to her
earnings in that bracket.

Figure~\ref{fig:2019taxbrackets} explains this rule in another way.

%Table~\ref{table:2019taxtable2} explains this rule in another way.

%\begin{table}
%\centering
%\begin{tabular}{
%rrrcl
%}
%\toprule
%\multicolumn{2}{c}{Taxable income} & & \\
%{more than} & {less than} & \multicolumn{3}{c}{Tax} \\
%\midrule
%\$0 & \$9,700 & && 10\% of taxable income \\
%\$9,700 & \$39,476 & \$970 & $+$ & 15\% of the excess over \$9,075 \\
%\$36,900 & \$89,350 & \$5,081 & $+$ & 25\% of the excess over \$39,600 \\
%\$89,350 & \$186,350 & \$18,194 & $+$ & 28\% of the excess over \$89,350 \\
%\$186,350 & \$405,100 & \$45,354 & $+$ & 33\% of the excess over \$186,350 \\
%\$405,100 & \$406,750 & \$117,541 & $+$ & 35\% of the excess over \$405,100 \\
%\$406,750 &  & \$118,189  & $+$ & 39.6\% of the excess over \$406,750 \\
%\bottomrule
%\end{tabular}
%\begin{csmr}[2014 single taxpayer tax calculation\label{table:2019taxtable2}]
%K. P. Erb,
%IRS Announces 2018 Tax Brackets, 
%Forbes.com (October 31, 2013),
%\url{www.forbes.com/sites/kellyphillipserb/2013/10/31/irs-announces-2014-tax-brackets-standard-deduction-amounts-and-more/}
%\access{August 12, 2015}.
%\csmrcomment{public tax bracket data}
%\end{csmr}
%\end{table}
%

\figfile{2019taxbrackets.pdf}
\begin{figure}
\centering
\includegraphics[width=\textwidth]{\thefigurefilename}
\begin{csmr}[2019 tax brackets\label{fig:2019taxbrackets}]
  K. Phillips Erb,
  IRS Announces 2019 Tax Rates, Standard Deduction Amounts And More,
Forbes, November 15, 2018,
\url{www.forbes.com/sites/kellyphillipserb/2018/11/15/irs-announces-2019-tax-rates-standard-deduction-amounts-and-more/\#44c55bc12081}
\access{October 21, 2020}.
\end{csmr}
\end{figure}
\figfile{}

\begin{teacher}
In the tax rate brouhaha in the 2008 election we recall reading a
story about a dentist who complained that he would need to be careful
not to let his income exceed \$250,000 --- where candidate Obama drew a
no-new-taxes line --- lest his overall tax rate increase. If you find
the story let us know and we'll turn it into an exercise.
\end{teacher}

The first graph in Figure~\ref{fig:GraduatedTax1019cropped}
from the spreadsheet \link{GraduatedTax2019.xlsx}
shows that the dependence of tax on income is \emindex{piecewise linear}
--- built from pieces of straight lines that become steeper as income
increases.  

\figfile{GraduatedTax2019cropped.pdf}
\begin{figure}
  \centering
  \includegraphics[width=\textwidth]{\thefigurefilename}
\caption{2019 single taxpayer tax liability}
\figsource{Charts from an Excel spreadsheet we built.}
\label{fig:GraduatedTax1019cropped}
\end{figure}
\figfile{}

The second graph shows the \myindex{effective tax rate}
--- the percentage of your income you pay in federal income tax. In
Equation~\ref{eq:fiftyK} we found that the total tax on a taxable income of
\$50,000 was  \$6858.50.
The effective tax rate is 
$\$6858.50 / \$50,000 = 13.72\%$. 
This is a weighted average of the three bracket rates
10\%, 12\% and 22\%, with weights the amount of income taxed at each rate.
The effective tax rate is less than the rate in your top bracket
because you pay at a
lower rate on the first part of your income.
The effective rate does not reach 35\% until about \$2 million in
income --- well into the top 37\% bracket.

In fact, the actual effective tax rate is lower than this for
wealthier households because income tax is collected only on 
income from wages and earnings. Income from capital gains ---
returns on investment --- is taxed at a lower
rate.
Figure~\ref{fig:FederalTaxRateByIncome2007} shows the actual effective
federal tax rate by total household income for the year 2007. 
The rate for the wealthiest households was just 16.6\% --- less than
half the 35\% rate for the top income tax bracket that year. It's
reasonable to assume that a similar discrepancy is still true.

\figfile{FederalTaxRateByIncome2007cropped.pdf}
\begin{figure}
\centering
\includegraphics[width=4in]{\thefigurefilename}
\begin{csmr}[Actual effective federal tax rate by income, 2007\label{fig:FederalTaxRateByIncome2007}]
Data from G. Anrig,
10 Reasons to Eliminate the Tax Break for Capital Gains,
The Century Foundation (October 20, 2011),
\url{tcf.org/blog/detail/10-reasons-to-eliminate-the-tax-break-for-capital-gains}
\access{August 4, 2015}.
\end{csmr}
\end{figure}
\figfile{}

\maasection{Social security}

Social security tax payroll deductions show up labelled
\index{FICA| see {Social Security}}
``FICA'' on your pay stub. That acronym is from
the ``Federal Insurance Contributions
Act''. Those taxes pay for \myindex{Social Security} and
\myindex{Medicare}.

In 2019 the starting tax rate was 6.2\% for Social Security and 1.45\% for
Medicare. The Social Security tax is collected only on
the first \$132,900 of your earnings. Up to that income level the
combined rate is 7.65\%.

When your earnings exceed $132,900$ you pay no more Social Security
tax, but you continue to pay Medicare tax at the 1.45\% rate. When
your income reaches \$200,000 the Medicare rate increase to 2.35\% on
the amount over \$200,000.

The actual rules are a little more
complicated. First, the tax applies only to wages. Other income (like
stock dividends or interest) are not subject to this tax. Second, the
real rates are twice the quoted amounts, but  your employer is
required to pay half. If you're self-employed you pay it all.


If you earn \$500,000 your FICA tax is 
%
\begin{align*}
  0.062 \times \$132,9000 & + 0.0145 \times \$200,000
   + 0.0235 \times (\$500,000 - \$200,000) \\
 & = \$18,190.
\end{align*}



Since you pay no Social Security on wages over \$132,900 
the percentage of your earnings collected for FICA taxes
decreases as your earnings increase even though you still pay for
Medicare. So 
FICA taxes are \emph{regressive}\index{regressive tax}.
Up to \$139,700 the effective rate is 7.45\%.
For \$500,000 the effective rate is just
$\$18,190/\$500,000 = 3.64\%$. For higher incomes, the effective rate
is even smaller. At huge incomes 
it levels off at the top Medicare rate of 2.35\%.
Figure~\ref{fig:socialsecurity} from spreadsheet 
\link{SocialSecurityTax2019.xlsx} shows the amount of FICA tax paid and
the decreasing effective tax rate as a function of FICA earnings.

\figfile{SocialSecurityTax2019cropped.pdf}
\begin{figure}
\centering
  \includegraphics[width=4in]{\thefigurefilename}
\caption{FICA (Social Security and Medicare) tax}
\label{fig:socialsecurity}
\end{figure}

\maasection{Income tax history}

For complex historical, legal and political reasons
the Supreme Court rejected the first attempts to collect an income
tax. This Constitutional amendendment, ratified in 1909, made that
kind of tax legal.

\begin{quotation}
Amendment XVI

The Congress shall have power to lay and collect taxes on incomes,
from whatever source derived, without apportionment among the several
States, and without regard to any census or enumeration.
\begin{csmr}
  United States of America 1789 (rev. 1992), 
  \url{www.constituteproject.org/constitution/United_States_of_America_1992}
  \access{September 25, 2019}.
\end{csmr}
\end{quotation}

The first federal income taxes were collected in 1913.
Figure~\ref{fig:toptaxrate} shows the significant fluctuations in the
rate for the top bracket through 2019. They have been near their
historic lows since the late 1980's. 

To assess the social and economic
impact of the changes in income tax rates would require much more time
and knowledge than we can offer here. 
The spreadsheet \link{Federalindividualratehistory.xlsx} contains a
complete history of income tax brackets and rates from the inception
of the income tax in 1913 through its hundredth anniversary in 2013,
in both dollars current in each year and adjusted for inflation (2012
dollars). 


\figfile{toptaxrate.pdf}
\begin{figure}
\centering
\includegraphics[width=\textwidth]{\thefigurefilename}
\begin{csmr}[Historical top tax bracket rate\label{fig:toptaxrate}]
  Bradford Tax Institute,
  History of Tax Rates: 1913 – 2019,
  \url{bradfordtaxinstitute.com/Free_Resources/Federal-Income-Tax-Rates.aspx}
  \access{February 2, 2019}.
\end{csmr}
\end{figure}
\figfile{}

Even in years when Congress does not revise the tax code, the
IRS routinely adjusts the brackets (not the rates) to take inflation
into account. If that were not done then salaries increased by
inflation would move people into higher brackets even when their
increased wages did not correspond to increased purchasing power.

Figure~\ref{fig:2018taxbrackets} shows the brackets for the tax year
2018. You can check that the 2019 brackets are all just about 2\%
larger than these.

\figfile{2018taxbrackets.pdf}
\begin{figure}
\centering
\includegraphics[width=\textwidth]{\thefigurefilename}
\begin{csmr}[2018 tax brackets\label{fig:2018taxbrackets}]
  K. Phillips Erb,
New: IRS Announces 2018 Tax Rates, Standard Deductions, Exemption
Amounts And More,
Forbes, March 7, 2018,
\url{www.forbes.com/sites/kellyphillipserb/2018/03/07/new-irs-announces-2018-tax-rates-standard-deductions-exemption-amounts-and-more/\#16f3fe9f3133}
\access{February 7, 2019}.
\end{csmr}
\end{figure}
\figfile{}

%Social security began in 1935 as a program to guarantee some financial
%security to older retired workers. It has grown over the
%years. Medicare was added in 2965.
%

%Figure 2019taxbrackets.png
%https://www.forbes.com/sites/kellyphillipserb/2018/03/07/new-irs-announces-2018-tax-rates-standard-deductions-exemption-amounts-and-more/#6bb2ab533133




\exstart

\begin{exx}{\untested\sref{tamworth}\gref{linearfunctions}
\gref{slopeintercept}}Your electricity bill.

Verify the computations on your current electricity bill, either with
a calculator or by modifying the Tamworth bill spreadsheet at
\link{TamworthElectric.xlsx}.

If you don't have a current electricity bill, check the website for
your local electric company, which \myindex{probably} provides a
sample bill you can use instead.

\end{exx}

\begin{exx}{\untested\sref{tamworth}\gref{adjustforinflation}
\gref{linearfunctions}}
Electricity costs now and then and here and there.

Compare residential electricity cost in Boston in 2007
(the date of the NStar bill in Figure~\ref{nstarbill}) to the cost
where you live today.  Your answer should take inflation into account.

If you have a current electricity bill, use it.
If you don't, try to find the fixed monthly cost and the cost of
electricity in \$/kwh from your local electric company. Perhaps their
website has that information. 

\end{exx}

\begin{exx}{\routine\hassolution\sref{tamworth}\gref{linearfunctions}
\gref{slopeintercept}}
How much electricity does it use \ldots ?

The document containing the Tamworth bill asks how 
much electricity various appliances use. 

\begin{enumerate}

\item Calculate how much electricity is consumed by a
\begin{abcd}

\item 	100w lamp on for 2 hours.
\item 500w TV on for 5 hours.
\item 2kw kettle on for half an hour.
\item 10w electric blanket on for 15 minutes.
\end{abcd}

\item Compare the electricity used by

\begin{abcd}
\item 2kw heater for 2 hours or 3 kw heater on for 3 hours.
\item 900w toaster for 15 minutes or 2kw grill for 10 minutes.
\item 100w radio for 2 hours or 500w radio for 45 minutes.
\item Which appliance would be cheaper to use in each case?
\end{abcd}
\end{enumerate}

\begin{sol}
\begin{enumerate}

\item Calculate how much electricity is consumed by a
\begin{abcd}

\item 	100w lamp on for 2 hours.
\begin{equation*}
100 \text{w} \times 2 \text{ hours} \times \frac{\text{kilo}}{1000} = 
0.2 \text{ kilowatt-hours}.
\end{equation*}

\item 500w TV on for 5 hours.

\begin{equation*}
500 \text{w} \times 5 \text{ hours} \times \frac{\text{kilo}}{1000} = 
2.5 \text{ kilowatt-hours}.
\end{equation*}

\item 2kw kettle on for half an hour.

1 kilowatt-hour. 

\item 10w electric blanket on for 15 minutes.
\begin{equation*}
10 \text{w} \times 0.25 \text{ hours} \times \frac{\text{kilo}}{1000} = 
0.0025 \text{ kilowatt-hours}.
\end{equation*}

\end{abcd}

\item Compare the electricity used by

\begin{abcd}
\item 2kw heater for 2 hours or 3 kw heater on for 3 hours.

Clearly 4kw hours is less than 9 kw hours. 

\item 900w toaster for 15 minutes or 2kw grill for 10 minutes.

It's easier to do this one if I leave the times in minutes.

$0.9 \times 15 = 1.35$ kw-minutes is less than
$2 \times 10 = 20$  kw-minutes.

\item 100w radio for 2 hours or 500w radio for 45 minutes.

I'll do this one in minutes too, and use watts rather than kilowatts.

$100 \times 120 = 12,000$ watt-minutes is less than 
$500 \times 45 = 22,500$ watt-minutes.


\item Which appliance would be cheaper to use in each case?

In each case the first one is cheaper - the amount of electricity is
about half.

\end{abcd}

\end{enumerate}
\end{sol}
\end{exx}

\begin{exx}{\hassolution\sref{tamworth}
\gref{directproportion}\gref{energyandpower}}
Direct current.

In an article from \theTimes{} on November 17, 2011 headlined
``From Edison's Trunk, Direct Current Gets Another Look''
\index{Edison, Thomas Alva}
you can read that

\begin{quotation}
In a data center redesigned to use more direct current,
  monthly utility bills can be cut by 10 to 20 percent, according to
  Trent Waterhouse, vice president of marketing for power electronics
  at General Electric. Verizon Communications, a G.E. customer,
  expects to save 1 billion kilowatt-hours a year from a nationwide
  retrofit of its data centers, which translates to roughly enough to
  power 77,000 homes.%
\begin{csmr}
M. Kanellos,
From Edison's Trunk, Direct Current Gets Another Look,
\theTimes{} (November 17, 2011),
\url{www.nytimes.com/2011/11/18/business/energy-environment/direct-current-technology-gets-another-look.html}
\access{July 22, 2015}.
\csmrcomment{68 words fair use}
\end{csmr}
\end{quotation}

\begin{abcd}
\item How many watt-hours is a billion kilowatt-hours?

\item About how much electricity is Verizon using now, given that they hope
to save a billion kilowatt-hours per year?

\item Verify the estimate that those billion kwh are ``roughly enough
to power 77,000 homes''. (You won't find anything useful in the
original article. How else will you search?)

\end{abcd}

\begin{sol}
\begin{abcd}
\item How many watt-hours is a billion kilowatt-hours?

Since ``kilo'' means ``multi
ply by 1000'', a billion kilowatt-hours is
(a billion times a thousand) watt-hours. That's a trillion
watt-hours, or $10^{12}$ watt-hours, or a terawatt-hour. If you have
to write the zeroes, it's 1,000,000,000,000 watt-hours.

\item About how much electricity is Verizon using now, given that they hope
to save a billion kilowatt-hours per year?

If the billion kwh saved were 10\% of what they are using now, that
would mean they were using 10 billion kwh now. If the billion kwh were
20\% of current use, that would make current use 5 billion kwh. So
they are using between 5 and 10 billion kwh now.

\item Verify the estimate that those billion kwh are ``roughly enough
to power 77,000 homes''. (You won't find anything useful in the
original article. How else will you search?)

The website 
\url{www.eia.gov/tools/faqs/faq.cfm?id=97&t=3}
says that

\begin{quotation}
In 2010, the average annual electricity consumption for a
U.S. residential utility customer was 11,496 kwh, an average of 958
kilowatthours (kwh) per month. Tennessee had the highest annual
consumption at 16,716 kwh and Maine the lowest at 6,252 kwh. 
\end{quotation}

Using the figures in the original quotation
%
\begin{equation*}
\frac{ 1 \text{ billion kwh}}{77,000 \text{ homes}}
= 12,987.013 
\frac{\text{kwh}}{\text{home}},
\end{equation*}
%
which is close enough to the government's figure of 11,496 kwh/home.

\end{abcd}

\end{sol}

\end{exx}

\begin{exx}{\hassolution\sref{linearfunction}\gref{linearfunctions}}
How hot was it?

Figure~\ref{fig:weather} shows a weather forecast for Hamilton,
Ontario, Canada. The temperature there is displayed in degrees Celsius,
marked ${}^{\circ}\text{C}$. 

\figfile{HamiltonWeather.pdf}
\begin{figure}
\centering
\framebox{
  \includegraphics[width=4in]{\thefigurefilename}
}
\caption{Weather in Canada}
\label{fig:weather}
\end{figure}
\figfile{}

\begin{abcd}
\item Look up the linear relationship for converting temperatures
measured on the Celsius scale to temperatures on the Fahrenheit scale.

\item Use the formula you found to calculate the 
temperature in degrees Fahrenheit in Hamilton, Ontario
on Thursday August 13, 2015 at 12:45.

\item Check your answer using the Google calculator.

\item What does ``Wind: SW 13 km/h'' in the figure mean?

\item What does ``Pressure: 101.8 kPa'' in the figure mean?

\item What is the \myindex{Perseid meteor shower}? 
\end{abcd}

\begin{sol}


\begin{abcd}
\item Look up the linear relationship for converting temperatures
measured on the Celsius scale to temperatures on the Fahrenheit scale.

The formula is 
%
\begin{equation*}
F = \frac{9}{5}C + 32 .
\end{equation*}

\item Use the formula you found to calculate the 
temperature in degrees Fahrenheit in Hamilton, Ontario
on Thursday August 13, 2015 at 12:45.

\begin{equation*}
\frac{9}{5}25 + 32 = 77.
\end{equation*}

\item Check your answer using the Google calculator.

I did. Same answer.

\item What does ``Wind: SW 13 km/h'' in the figure mean?

Wind blowing from the southwest at 13 kilometers per hour.

\item What does ``Pressure: 101.8 kPa'' in the figure mean?

It's a measure of air pressure. ``Pa'' is a ``Pascal'', a unit of
pressure. The k is kilo, of course. So the pressure was 101.8 thousand
Pascals.

\item What is the \myindex{Perseid meteor shower}? 

Every year in August the Earth passes through a cloud of rocky junk in
space that leads ot a lot of meteors as the junk burns up in the
atmosphere. 

\end{abcd}

\end{sol}

\end{exx}

\begin{exx}{\hassolution\sref{linearfunction}\gref{linearfunctions}
\gref{slopeintercept}\gref{functions}} 
The \myindex{Jollity Building}.

In \exref{liebling} there's an implicit linear model
for Mr. Ormont's weekly income. Write the linear equation for that
model. Clearly identify the independent and dependent variables and
the units for the slope and the intercept.

\begin{sol}
The clue to finding the linear equation is the phrase
``fifty dollars a week and 2 percent of the total rents.'' I'll let
$W$ stand for Mr. Ormont's weekly income and $R$ for the weekly rent.
Then 
\begin{equation*}	
	W = \$50 + 0.02R.
\end{equation*}
The constant amount (the intercept) is \$50. He gets that much even if
the building is empty. The slope is 0.02. Its units are
\begin{equation*}
\frac{\text{dollars for Mr. Ormont}}
{\text{dollar of weekly rent}}.
\end{equation*}

\end{sol}
\end{exx}

\begin{exx}{\hassolution\sref{linearfunction}
\gref{linearfunctions}\gref{slopeintercept}}
Newton trees .

In an article in the April 2012 issue of the \myindex{Newton
Conservators} newsletter you can read that

\begin{quotation}
In the early 1970s there were approximately 40,000 trees
lining the streets of Newton. Today, that number is about 26,000 --- a
35\% loss. The current annual rate of decline is about 650 trees per
year. At this rate, if unchecked, public street trees would diminish to 
approximately 10,000 within a generation (25 years), and in 40 years,
public street trees would no longer be part of the Newton landscape.%
\begin{csmr}
From ``Public Street Trees – A Choice'' by J. DiMiceli in Newton
Conservators Newsletter (April 2012),
\url{newtonconservators.org/public-street-trees-a-choice/}
\access{July 28, 2019},
Quoted with permission.
\end{csmr}
\end{quotation}

\begin{teacher}
Exponential decay might be a better model than linear for this
exercise on the Newton trees. Consider returning to it when
you reach that topic later in the text. 
\end{teacher}

\begin{abcd}
\item Check the arithmetic that leads to the claimed ``35\% loss''.

\item Check the arithmetic that leads to a ``current annual rate of
650 trees per year''.

\item Check the predictions in the last sentence. Are they 
likely\index{probability}  to
come to pass?

\item Write the equation  for the linear model implicit in this
quotation (use years since 2012 as the independent variable).
Identify the slope and the intercept, with their units.%

\end{abcd}

\begin{sol}

\begin{abcd}
\item Check the arithmetic that leads to the claimed ``35\% loss''.

The 1+ trick tells me $40\text{K} \ times 0.65 = 26\text{K}$.

\item Check the arithmetic that leads to a `` current annual rate of
650 trees per year''.

If I take ``the early 1970s'' to be 1972 then the interval is 40 years.
\begin{equation*}
\frac{14,000 \text{ trees}}{40 \text{ years}} =
350 \frac{\text{trees}}{\text{year}}.
\end{equation*}
That doesn't match the figure in the article. Perhaps the rate of loss
is larger in more recent years.

\item Check the predictions in the last sentence. Are they 
likely to come to pass?

Using the rate of 650 trees per year, the loss in 25 years would be
$25 \times 650 \approx 16,000$ trees. Subtracting that from the
current 26,000 trees would indeed leave just 10,000.

In another 15 years the loss would be just about all the 10,000
remaining trees.

\item Write the equation  for the linear model implicit in this
quotation (use years since 2012 as the independent variable).
Identify the slope and the intercept, with their units.

The independent variable is $Y$, years since 2012. The dependent
variable is $T$, the number of trees. The equation is
\begin{equation*}
T = 26,000 - 650Y.
\end{equation*}
The intercept is 26,000 trees, the slope is $-650$ trees/year.
\end{abcd}

\end{sol}
\end{exx}

\begin{exx}{\routine\hassolution\artificial\sref{linearexcel}
\gref{linearfunctions}\gref{slopeintercept}\gref{linearexcel}}
Apple time.

Kuipers Family Farm (\url{www.kuipersfamilyfarm.com/apple-orchard/}),
in Maple Park, IL, charged \$10 per person for 
admission to the apple orchard during the 2018  apple season.  This included a
$\frac{1}{2}$ peck bag of apples and a hayride to the orchard.
Visitors could pick additional peck bags of apples for \$15 each. (A 
bag of apples in the grocery store usually contains $\frac12$ peck.)  
 
\begin{abcd} 	
\item  If you pick that additional peck of apples, how much do you pay?
\item  If you pick two additional pecks of apples, how much do you pay?
\item  If you just enjoy the free cider, your $\frac12$ peck of apples
  and the sunshine, what do you pay? 
\item  Write a linear function that shows how the total cost of the
farm visit depends on the how many pecks of apples you pick. Identify
the slope and intercept, with their units. 
\item Build an Excel spreadsheet to compute this function and use it
to check the values you worked out by hand above. Include a chart in
your spreadsheet.
\item Use your spreadsheet to calculate your apple cost (in
  \$/peck) when you pick no extra pecks, then when you pick 1, 2 or
  10 extra pecks.
\end{abcd}

\begin{sol}

I found the function first. If $C$ is the total cost of the visit when
you pick $A$ pecks of apples then $C$ depends on $A$ this way:
%
\begin{equation*}
	C = 15 A + 10.00.
\end{equation*}
%
That's a linear function with slope 15~\$/peck and intercept
\$10.00.

I used the function to find the  following answers.

\begin{abcd} 	
\item If you pick that additional peck of apples, how much do you pay?

\$25.00.

\item   If you pick two additional pecks of apples, how much do you pay?

\$40.00.

\item   If you just enjoy the free cider, your $\frac12$ peck of
  apples and the sunshine, what do you pay? 


\$19.00. Didn't need the formula for this one.

\item  Write a linear function that shows how the total cost of the
farm visit depends on the amount of apples picked. Identify
the slope and intercept, with their units.

Did that first.

\item Build an Excel spreadsheet to compute this function and use it
to check the values you worked out by hand above.

See \slink{U-PickApplesSolution.xlsx}

\item Use your spreadsheet to calculate your apple cost (in
  \$/peck) when you pick no extra pecks, then when you pick 1, 2 or
  10 extra pecks.

The number of pecks of apples you go home with is $A + \frac12$. I
used the spreadsheet to compute $C/(A+ \frac12)$ for $A=0, 1, 2$ and
$10$ and found the cost per peck to be $\$20.00, \$16.67, \$16.00$ and
$\$15.24$. I'm not surprised that the more you pick the closer the
cost per peck is to $\$15$.

\end{abcd}

\end{sol}
\end{exx}

\begin{exx}{\hassolution}{\artificial\sref{comparelinear}
\gref{linearfunctions}\gref{linearexcel}} 
Comparing telephone calling plans.\index{telephone bill} 

A cell phone company has introduced a pay-as-you-go price structure,
with three possibilities.

\begin{center}
\begin{tabular}{ccr}
\toprule
	Plan 1 &  \$10 a month & 10 cents per minute \\
	Plan 2 &  \$15 a month  & 7.5 cents per minute \\
	Plan 3 &  \$30 a month & 5 cents per minute \\
\bottomrule
\end{tabular}
\end{center}

\begin{abcd}

\item For each plan, find a linear function that describes
how the total cost for one month depends on the number of minutes
used. 

\item Construct a table in Excel showing the total cost for
one month for each of the three plans. Organize your data this way:

\begin{itemize}
\item Create a sequence of cells in column A for the various possible
numbers of minutes. Label that column. Start with 0 minutes. What's a
good step to use? What's a reasonable place to stop?

\item Use columns \cell{B}, \cell{C} and \cell{D} for each of the
  three plans. The fixed charge  and charge per call should be in
  cells in those columns too, so you can use the same formula
  everywhere in the data table. (That will call for clever use of
  the \excel{\$} to keep Excel from changing row numbers and column
  letters when you don't want it to.)  
\end{itemize}

\item Use Excel to draw one chart showing how the monthly bill
  ($y$-axis) depends on the number of minutes you use the phone
  ($x$-axis) for all three plans.  

\item Write a paragraph explaining to your friend how she should go about
choosing the plan that's best for her. 
\end{abcd}

\begin{hint}
Finding the places
where the lines in your Excel chart cross is the key to the last part
of the problem.
\end{hint}


\begin{sol}

\begin{abcd}

\item For each plan, find a linear function that describes
how the total cost for one month depends on the number of minutes
used. 

Let $B$ represent the monthly cost, in dollars, and $M$ the
number of minutes used. Then the three linear functions are
\begin{align*}
	B & =  0.10 M + 10, \\
	B & =  0.075 M + 15, \\
	B & =  0.05 M + 30.
\end{align*}

\item Use Excel to draw one chart showing how the monthly bill
  ($y$-axis) depends on the number of minutes you use the phone
  ($x$-axis) for all three plans.  

See the spreadsheet at \slink{PhoneCostsSolution.xlsx}.

\item 
If you talk for less than 200 minutes, choose Plan 1. If you talk for
more than 600, choose Plan 3. For anything in between, choose Plan
2. You can check my work by looking at the spreadsheet --- the
conclusion is plain from both the table and the chart.

\end{abcd}

\end{sol}

\end{exx}

\begin{exx}{\hassolution\sref{comparelinear}\gref{linearfunctions}
\gref{linearexcel}}
Prepaid phones\index{prepaid phones}.

Until summer 2012, if you wanted an \myindex{iPhone} you needed to
lock into a two-year contract.  Then some mobile companies started
selling the iPhone and letting you choose your own plan, with no contract.
You can use the ideas from this chapter to make a quantitative
comparison (we'll let you decide what other factors, such as paying a
lot up front, matter in your decision).

Virgin Mobile began selling a 16GB iPhone 4S for
\$649.99, with no plan or contract.  They offered a \$55 per month
``unlimited" data plan (in fact it was not unlimited, as once you go
past 2.5GB of data they slowed the phone speed down considerably).  If
you purchased the phone through Sprint, it cost just \$149.99.
However, you had to sign up for a two-year contract.  The least
expensive option was \$79.99 with what they called unlimited data. 

\begin{abcd*}
\item  For each plan, find a linear function that shows how the total cost of
the phone depends on the number of months you have it.

\item  Build a spreadsheet in Excel using your functions and fill in
  the cost of the two different options over several months. 

\item  Graph the data in your spreadsheet.  

\item  Write a short statement comparing the two plans.  Clearly the
  Virgin Mobile plan is more expensive at first.  When does it become
  the less expensive plan?  If you had to choose one of the plans,
  which one would you choose and why?  What other factors would you
  consider? 
\end{abcd*}

\begin{sol}
The answers to all the questions are in the spreadsheet at
\slink{PrepaidPhonesSolution.xlsx}.

\end{sol}
\end{exx}


\begin{exx}{\hassolution\sref{comparelinear}\gref{slopeintercept}
\gref{linearexcel}}
Hybrid payback.\index{hybrid}

The ``Best \& Worst Cars 2011'' issue of \emph{Consumer Reports}
provides the following data for new Toyota Camrys:

\begin{center}
\begin{tabular}{ccc}
 & conventional & hybrid \\
\midrule
cost & \$19,720 & \$26,575 \\
fuel economy & 26 MPG & 34 MPG \\
\bottomrule
\end{tabular}
\end{center}

Assume gasoline costs \$3.50/gallon.

\begin{abcd}
\item Questions about the conventional Camry. 

\begin{enumerate}[(i)]

\item
Once you own the car, how much does it cost to run, in 
dollars per mile? Does your answer make sense?

\item
Calculate the total cost (purchase plus gasoline) to drive the
conventional Camry 10,000 miles.

\item
Write the linear equation that computes the total cost $C$ of
driving the conventional Camry $M$ miles.

\item
Identify the slope and the intercept of this equation, with
their units. 
\end{enumerate}

\item Open the spreadsheet \link{ConventionalvsHybrid.xlsx}. Enter the
numerical data from the table and the cost of
gasoline in the appropriate cells. What formula should you enter in cell
\cell{C15} to check your answer to part (i)? 

\item Copy your formula to cells \cell{B14:D29} to fill in the
table. Where must you add \$ signs to keep Excel from
changing row and column references?

\suspend{abcd}

Create a properly formatted and labelled chart in Excel showing how
the cost of driving each car depends on the number of miles
driven. Use your graph along with the table to answer the following
questions.

\resume{abcd}

\item If you drive 120,000 miles will you recover in gas savings the
extra initial cost of the hybrid? Write a complete sentence or two and
use appropriate precision for the numbers you use to make your
argument.

\item When will you recover the extra initial cost in gas savings 
if the government (re)instates a \$3,000 tax rebate for hybrid
purchases? 

\item With the original initial costs, how much would the price of
gasoline have to be in order for the breakeven point to occur at
30,000 miles?

\item Restore all the inputs to their original values. Arrange 
  your spreadsheet so that it will print on one page, with the chart
  below the data table.

\end{abcd}

(If you're thinking of buying a car, remember that there's
a lot more that goes into the cost of driving one car or
another (or any car at all) than just these simple computations using
initial cost and miles driven.)

\begin{sol}

\begin{abcd}
\item Questions about the conventional Camry. 

\begin{enumerate}[(i)]

\item Once you own the car, how much does it cost to run, in 
dollars per mile? Does your answer make sense?

The cost in dollars per mile is
\begin{equation*}
3.50 \frac{\$}{\text{gallon}} \times 
\frac{ 1 \text{ gallon}}{26 \text{ miles}}
= 0.1346 \frac{\$}{\text{mile}}.
\end{equation*}

That's about 13 cents per mile, which makes sense to me.

\item Calculate the total cost (purchase plus gasoline) to drive the
conventional Camry 10,000 miles.

\begin{equation*}
\$19,720 + 
0.1346 \frac{\$}{\text{mile}} \times
10,000 \text{ miles} 
=\$21,066.1538 \approx \$21,000.
\end{equation*}

\item Write the linear equation that computes the total cost $C$ of
driving the conventional Camry $M$ miles if gasoline costs
\$3.50/gallon.
\begin{equation*}
	C = 19,720 + 0.1346\ M.
\end{equation*}

\item Identify the slope and the intercept of this equation, with
their units. 

The slope is 0.1346 dollars per mile. The intercept
is 19,720 dollars.
\end{enumerate}

\item Open the spreadsheet \link{ConventionalvsHybrid.xlsx}. Enter the
data from the table on page 1 and the cost of gasoline in the
appropriate cells. What formula should you enter in cell
\cell{C15} to check your answer to part (i) ? 

\displayexcel{
=C9+C6*B15/C10 
}

\item Copy your formula to cells \cell{B14:D29} to fill in the
table. Indicate here where you put \$ signs in to keep Excel from
changing row and column references.

\displayexcel{
=C\$9+\$C\$6*\$B15/C\$10 
}

Note that I have no \cell{\$} before two of the \excel{C} references so this
formula will do the right thing when I move it to column \cell{D} for
the hybrid.

\suspend{abcd}

Create a properly formatted and labelled chart in Excel showing how
the cost of driving each car depends on the number of miles
driven. Use your graph along with the table to answer the following
questions.

See \slink{ConventionalvsHybridSolution.xlsx}. 

\resume{abcd}

\item If you drive 120,000 miles will you recover in gas savings the
extra initial cost of the hybrid? Write a complete sentence or two and
use appropriate precision for the numbers you use to make your
argument.

No. The total cost for driving the hybrid is about \$39K. That's about
\$3,000 more than the total cost for driving the conventional Camry.

\item When will you recover the extra initial cost in gas savings 
if the government (re)instates a \$3,000 tax rebate for hybrid
purchases? 

The breakeven point is at about 120,000 miles.
According to Excel the costs for the conventional and hybrid are
\$35,873.85 and \$35,927.94 respectively.  These are only about \$50
apart, which is much too much precision for a prediction that far
away.

\item With the original initial costs, how much would the price of
gasoline have to be in order for the breakeven point to occur at
30,000 miles?

By trial and error, changing the gasoline price in the spreadsheet, I
found out that gas would have to cost about \$23 per gallon for
this to happen. 

\end{abcd}

\end{sol}

\end{exx}


\begin{exx}{\hassolution\sref{comparelinear}\gref{slopeintercept}
\gref{linearexcel}}
Contract or not?

Table~\ref{table:cellphoneplans} provides data that appeared in a
story in \theGlobe{} on June 14, 2012 
headlined ``Pay full price for \myindex{iPhone}, avoid contract''.%

\begin{table}
\centering
\begin{tabular}{lS[table-format=3.2]S[table-format=2.0]S[table-format=4.2]}
\toprule
Plan & {Phone cost} & {Monthly charge} & {Two year cost} \\
\midrule
Cricket Wireless & 499.99 & 55 & 1,819.99 \\
Virgin Mobile &  649 & 30 & 1,369 \\
\bottomrule
\end{tabular}
\begin{csmr}[Comparing cell phone plans\label{table:cellphoneplans}]
Data from H. Bray,
Pay full price for iPhone, avoid contract,
\theGlobe{} (June 14, 2012),
\url{bostonglobe.com/business/2012/06/14/bgcom-techlab/AskcWIPBv1qccvmqIx7DmK/story.html}
(broken link) 
\access{July 22, 2015}.
\end{csmr}
\end{table}

\begin{abcd}

\item How much would it cost (in total) to buy the Cricket phone and
use it for two months?

\item Write an equation for the total cost to buy and use the
Virgin Mobile phone for $M$ months.

\item Identify the slope and the intercept of your equation, with
proper units for each. 

\item Create an Excel spreadsheet and use Excel formulas to complete a
table like this:

\begin{center}
\begin{tabular}{|c|c|c|}
\hline
Months & Cricket & Virgin \\
\hline
0 & & \\
1 & & \\
\dots & & \\
24 & & \\
\hline
\end{tabular} .
\end{center}

\item Check that your spreadsheet produces the answers in the table
for 24 months.

\item Create a properly labelled and formatted chart displaying the
data in your table. 

\item When (in terms of months of use) would it be better to choose
the Cricket phone?

\item Suppose the monthly charge for the Cricket phone was just \$45
while that for the Virgin phone increased to \$35/month. Answer the
previous question with this new data.

\end{abcd}

\begin{sol}
\begin{abcd}

\item How much would it cost (in total) to buy the Cricket phone and
use it for two months?

Cricket phone two months: \$609.99.

\item Write an equation for the total cost to buy and use the
Virgin Mobile phone for $M$ months.
%
\begin{equation*}
\text{Virgin Cost} = 650 + 30M.
\end{equation*}

\item Identify the slope and the intercept of your equation, with
proper units for each. 

The slope is 30 dollars/month; the intercept is 650 dollars.

\item Create an Excel spreadsheet and use Excel formulas to complete
  the table.

See \slink{cellphoneplansolution.xlsx}.

\item Check that your spreadsheet produces the answers in the graphic
for 24 months.

It does.

\item Create a properly labelled and formatted chart displaying the
data in your table. 

See \slink{cellphoneplansolution.xlsx}.

\item When (in terms of months of use) would it be better to choose
the Cricket phone?

Cricket is cheaper up to 6 months. Then Virgin is the better buy.

\item Suppose the monthly charge for the Cricket phone was just \$45
while that for the Virgin phone increased to \$35/month. Answer the
previous question with this new data.

Change values in cells \cell{C9} and \cell{D9}. Then the graphs cross at
14 months.

\end{abcd}

\end{sol}

\end{exx}

\begin{exx}{\hassolution\sref{energypower}\gref{linearfunctions}
\gref{directproportion}\gref{energyandpower}}
Regenerative braking.
\index{regenerative braking}

When you apply the brakes in a Toyota Prius 
\index{Prius} the car uses some of
the energy of the forward motion to recharge the battery.
The dashboard displays a little car icon each time
that recharging has collected 50 watt-hours. 

\begin{abcd*}
\item Estimate the energy equivalent of each icon in gallons of gasoline.
\item Estimate the dollar value of that gasoline.
\item Compare your estimate to the dollar value of 50 watt-hours of
electricity in your house (what it would cost to keep a 100 watt bulb
on for half an hour).
\item Discuss the value of the display.
\end{abcd*}

\begin{hint}
For part (a) 
you will have to look up the conversion factors among various forms of
energy in order to convert watt-hours to gallons of gasoline. Part (d)
is about psychology, not quantitative reasoning. Your answer might
begin ``The display is valuable because \ldots On the other hand \ldots''
\end{hint}

\begin{sol}

\begin{abcd*}
\item Estimate the energy equivalent of each icon in gallons of
gasoline.

\url{www.calculateme.com/energy/kilowatt-hours/to-gallons-of-gas/}
I discovered that
0.05 kilowatt-hour equals 0.0015 gallons of gasoline
so each icon is equivalent to between one  and two tenths of a percent
of a gallon of gasoline. 

\item Estimate the dollar value of that energy.

I'll use \$3 per gallon for the price of gasoline. Then 0.0015 gallons
will cost about 0.45 cents. Call it half a penny.

That was a much smaller number than I expected.

\item Compare your estimate to the dollar value of 50 watt-hours of
electricity in your house (what it would cost to keep a 100 watt bulb
on for half an hour).

Electricity in my house costs about 10 cents per
kilowatt-hour. So there too 50 watt-hours does indeed cost half a penny!
So I think my answer to the previous part of the problem is probably
right, even though it's surprising.

\item Discuss the value of the display.

Regenerative braking doesn't save much gasoline money at all ---
just pennies for even moderately long trips. So the value of the gas
is negligible. But the display might be valuable. I know from driving
my daughter's Prius on occasion and watching others drive that the
display of the energy efficiency encourages people to drive more
slowly and use less gas. 
\end{abcd*}

\end{sol}

\end{exx}

\begin{exx}{\hassolution\sref{energypower}
\gref{directproportion}\gref{energyandpower}}
Computers don't sleep soundly.

The website 
\url{michaelbluejay.com/electricity/computers.html} gives
information about how much energy a computer uses while asleep, in
standby mode, or in use.  The iMac G5, for example, uses 97 watts
while ``doing nothing'', compared to 3.5 watts while asleep.   
\begin{abcd}

\item  What do you think ``doing nothing'' means?

\item  If this type of computer is doing nothing all day (24 hours),
how much electricity does it use?  Express your answer in
kilowatt-hours.

\item  Now suppose the computer goes to sleep after 15 minutes of
doing nothing. How much electricity does it use in an idle day?

\item If a kilowatt-hour of electricity costs 20 cents, how much
money is saved in a day because the computer is smart enough to go to
sleep?

\end{abcd}

\begin{sol}

\begin{abcd}
\item  What do you think ``doing nothing'' means?

According to the quote, ``doing nothing'' can't really be doing
nothing, since it uses 97 watts. That's probably running the screen
saver program, and checking from time to time for email or updates to
various websites --- things the computer does when it's turned on even
if no one is typing at the keyboard or using the mouse.

\item  If this type of computer is doing nothing all day (24 hours),
how much electricity does it use?  Express your answer in
kilowatt\-hours.

Consuming electricity at a rate of 97 watts for a day uses
97 watt-days. That's
$24 \times 97 =2,328$ watt-hours or 2.3kwh of electricity.

\item  Now suppose the computer goes to sleep after 15 minutes of
doing nothing. How much electricity does it use in an idle day?

A lot less: $0.25 \times 97 + 23.75 \times 3.5= 107.375$ watt-hours,
which is about 0.11 kwh.

\item If a kilowatt-hour of electricity costs 20 cents, how much
money is saved in a day because the computer is smart enough to go to
sleep?

The sleep saves about 2.2 kwh of electricity, worth 44 cents.

\end{abcd}

\end{sol}
\end{exx}


\begin{exx}{\hassolution\complex\sref{energypower}
\gref{directproportion}\gref{energyandpower}}
How Much Water Does Pasta Really Need?

On February 24, 2009 \thetimes{} published an article
by Harold McGee addressing that question.
\index{pasta}

McGee's kitchen experiments convinced him that he could cook pasta in
far less water than is customary. Since (he says) we consume about a
billion pounds of pasta a year:

\begin{quotation}
My rough figuring indicates an energy savings at the stove top of
several trillion B.T.U.s. At the power plant, that would mean saving
250,000 to 500,000 barrels of oil, or \$10 million to \$20 million at
current prices. Significant numbers, though these days they sound like
small drops in a very large pot.%
\begin{csmr}
H. McGee,
 How Much Water Does Pasta Really Need?,
\theTimes{} (February 24, 2009),
\url{www.nytimes.com/2009/02/25/dining/25curi.html}
\access{July 22, 2015}.
\csmrcomment{58 words fair use}
\end{csmr}
\end{quotation}

\begin{abcd}

\item
Verify the author's conversion of ``several trillion B.T.U.s'' to
barrels of oil and then to dollars.

\item Does McGee's estimate of a billion pounds of pasta per year make
  sense?

\item How much water do Americans use cooking pasta? How much would
  they use if they followed McGee's advice?

\item Does not boiling the extra water really save the amount of energy
  McGee claims?
\end{abcd}


\begin{sol}

\begin{abcd}

\item
Verify the author's conversion of ``several trillion B.T.U.s'' to
barrels of oil and then to dollars.

The website
\url{www.unitjuggler.com/convert-energy-from-boe-to-MMBtu.html}
says that the energy in one barrel of oil is 5.55 million Btu.
If I take ``several trillion Btu'' to be ``2 trillion Btu'' then
that's equivalent to (2 trillion)/(5.5 million) = 360 thousand barrels
of oil. That's just about what McGee says. 

On October 24, 2014 oil cost about \$80/barrel. 250,000 barrels
would cost about \$20 million. That's double McGee's estimate, so when
he wrote in February 2009 oil must have cost about \$40/barrel. The web site
\url{www.macrotrends.net/1369/crude-oil-price-history-chart}
shows a cost of about \$42/barrel for January 2009.

\item Does McGee's estimate of a billion pounds of pasta per year make
  sense?

That would be about 
%
\begin{equation*}
\frac{1 \text{ billion pounds}}{100 \text{ million households}}
= 
10 \frac{\text{pounds}}{\text{household}} .
\end{equation*}
%
I guess I can believe that order of magnitude --- it's pasta for dinner
about once a month on average. Some households never cook it; those
that do probably have it much more often than that.

\item How much water do Americans use cooking pasta? How much would
  they use if they followed McGee's advice?

If the usual method is to use two quarts of water for a pound, this much
pasta would take two billion quarts of water. McGee suggests two cups 
instead of two quarts. Since there are four quarts in a cup, doing
pasta McGee's way would take only one fourth as much water, or half a
billion quarts.

\item Does not boiling the extra water really save the amount of energy
  McGee claims?

To answer this I looked up how much energy it takes to boil
water. Several websites told me it was about 350 kilojoules per
quart. Google told me that was about 330 Btu per quart. To boil the
extra billion and a half quarts of water would take about 500 trillion
Btu. That's more than the ``several trillion'' McGee claims. I wonder
where I went wrong --- if I did.


\end{abcd}

\end{sol}
\end{exx}


\begin{exx}[solarenergy1]{\hassolution\sref{energypower}\gref{energyandpower}}
  Solar energy.
\index{solar energy}

On May 1, 2013, \emph{The Arizona Republic} reported on 
the \$500 million Arlington Valley Solar Energy II project
near Phoenix.  The article said 

\begin{quotation}
[the project] will have 127 megawatts of capacity when finished. One
megawatt is enough electricity to supply about 250 Arizona homes at
once, when the sun is shining on the solar panels.%
\begin{csmr}
R. Randazzo,
New Arlington Valley solar site packs power,
\emph{The Arizona Republic} (May 1, 2013).
\url{www.azcentral.com/business/arizonaeconomy/articles/20130501new-arlington-valley-solar-site-packs-power.html}
\access{July 22, 2015}.
\csmrcomment{30 words fair use}
\end{csmr}
\end{quotation}

\begin{abcd}

\item How many watts of power does the average Arizona home need to
  run its appliances when the sun is shining?

\item Research the power requirements of several typical home
  appliances: air conditioners, stoves, television sets, \ldots
Then decide whether the article's claim about home power requirement
on a sunny day in Arizona is reasonable.

\end{abcd}


\begin{sol}

\begin{abcd}

\item How many watts of power does the average Arizona home need to
  run its appliances when the sun is shining?

Since one megawatt powers 250 homes, one home requires
%
\begin{equation*}
\frac{1}{250}\text{ mw} = 
\frac{1000}{250}\text{ kw} = 
4 \text{ kw} = 
\end{equation*}

\item Research the power requirements of several typical home
  appliances: air conditioners, stoves, television sets, \ldots
Then decide whether the article's claim about home power requirement
on a sunny day in Arizona is reasonable.

The website
\url{www.wholesalesolar.com/solar-information/how-to-save-energy/power-table}
lists the power requirements for many devices. I found out there that
central air conditioning can call for 5000 watts (1000 watts for room units).
A TV might need 200w. Add in the lights and computers, a toaster or
microwave or electric stove, electric water heat and other appliances
and I think 4 kilowatts is probably a low estimate.

\end{abcd}


\end{sol}
\end{exx}


\begin{exx}[solarenergy2]{\hassolution\complex\sref{energypower}
\gref{directproportion}\gref{energyandpower}}
Solar power at \myindex{Wellesley College}.

The sign on a solar panel array at Wellesley College reads:
\begin{center}
\framebox{
\begin{minipage}{4.5in}
\begin{center}
Solar Photovoltaic Array \\
This 10-kilowatt Solar PV Array is composed \\
of 48 panels, each 210 watts. It will generate approximately \\
13,000 kilowatt hours of electricity per year, enough to \\
power 2 homes, 32 metal halide street lights \\
or 85 LED street lights for an entire year. \\
For real time electrical output please go to: \\
\url{www.sunwatchmeter.com/home/day/wellesley-college} \\
PLEASE KEEP OFF THE PANELS
\end{center}
\end{minipage}
}
\end{center}

You can see a picture at
\url{www.theswellesleyreport.com/2010/09/wellesley-college-saves-the-planet/solar-panel/}

\begin{abcd}
\item Check the consistency of some of the numbers.

\item How many hours of sunshine per day do the designers expect the
installation to see?

\item Visit the website on the sign and write about it. What do the
graph and the meters represent? What is happening there now? 
\end{abcd}

\begin{sol}

\begin{abcd}
\item Check the consistency of some of the numbers.

The total power is $48 \times 210 \approx 50 \times 200 = 10,000$
watts, just at the sign says. (The exact arithmetic comes to 
$48 \times 210 = 10,080$ watts.)

\item How many hours of sunshine per day do the designers expect the
installation to see?

13,000 kilowatt-hours per year is $13,000 / 365 \approx 36$
kilowatt-hours per day. The 10,000 kilowatt power supply would take
3.6 hours of sunshine to generate than much electricity.

Three and a half hours of sunshine a day sounds like a reasonable
average. 

\item Visit the website on the sign and write about it. What do the
graph and the meters represent? What is happening there now? 

I visited
The graph shows time of day on the
$x$-axis and power (in watts) on the $y$-axis. The green area under
that graph is the number of watt-hours of electric energy generated so
far in the day. The first meter shows the power of the array at the
moment. The meter on the right shows whether Wellesley is generating
more or less electricity than it is using at the moment.

October 28, 2015, at 2 p.m. the power was 849 watts (less than a
kilowatt) and Wellesley was getting electricity from the grid.

\end{abcd}

\end{sol}
\end{exx}


\begin{exx}{\complex\sref{energypower}\gref{energyandpower}}
Wind power.
\index{wind power}

From \emph{The Los Angeles Times}, March 1, 2009:

\begin{quotation}
The U. S. last year surpassed Germany as the world's No.
1 wind-powered nation, with more than 25,000 megawatts in
place. Wind could supply 20\% of America's electricity needs by
2030, up from less than 1\% now, according to a recent Energy
Department report.%
\begin{csmr}
M. Dickerson,
Wind-power industry seeks trained workforce,
\emph{Los Angeles Times} (March 1, 2009),
\url{www.latimes.com/archives/la-xpm-2009-mar-01-fi-wind-bootcamp1-story.html}
\access{July 28, 2019}.
\csmrcomment{46 words fair use}
\end{csmr}
\end{quotation}

\begin{abcd}

\item What do these data say when you calculate wind power
in megawatts per person, or as a percentage of the total power
available? 

\item Explain why it might or might not be true
to say that when this article appeared the U.S. now produced more
wind energy than Germany?
\end{abcd}

\end{exx}

\begin{exx}{\hassolution\sref{energypower}\gref{energyandpower}
\gref{directproportion}}
World \myindex{solar power}.

In a posting on their website on July 31, 2013, the Earth Policy
Institute reported that

\begin{quotation}
The world installed 31,100 megawatts of solar photovoltaics
  (PV) in 2012 --- an all-time annual high that pushed global PV
capacity above 100,000 megawatts.  There is now enough PV operating to
meet the household electricity needs of nearly 70 million people at
the European level of use.%
\begin{csmr}
J. M. Roney,
World Solar Power Topped 100,000 Megawatts in 2012,
Earth Policy Institute (July 31, 2013),
\url{www.earth-policy.org/indicators/C47/solar_power_2013}
\access{July 23, 2015}.
\csmrcomment{49 words fair use}
\end{csmr}
\end{quotation}

A graphic at the website
\url{www.wec-indicators.enerdata.eu/household-electricity-use.html}
shows that average household electricity consumption in Europe in 2013
was about 4,000 kwh/year.%
\begin{csmr}
Average electricity consumption per electrified household,
\url{www.wec-indicators.enerdata.eu/household-electricity-use.html},
\access{November 14 2015}.
\end{csmr}

\begin{abcd}

\item  Use the data in this exercise to estimate the average number of
  hours per day that these solar panels are producing electricity.

\item  How does average household electricity consumption in the
  U.S. compare to that in Europe? 

\begin{hint}
It should not be hard to find a website that helps with the second
question.
\end{hint}


\end{abcd}


\begin{sol}
\begin{abcd}

\item  Use the data in this exercise to estimate the average number of
  hours per day that these solar panels are producing electricity.


  The quote says that 100,000 megawatts is power for 70 million
  people. If I estimate three people per household on average that's
  about 25 million households, which is 25 megahouseholds. That comes
  to about 4,000 watts, or 4 kilowatts per household.

  Since European households use 4,000 kilowatt-hours per year, that
  solar panels would have to be working 1,000 hours per year to
  generate that electricity --- about one eighth of the 8,000 hours
  per year. So the panels would be working about 3 hours per day.

\item  How does average household electricity consumption in the
  U.S. compare to that in Europe? 

The same website that provided European household use shows shows a
value of about 12,000 kwh/year for the United States. That's about
three times the European average.

\end{abcd}

\end{sol}

\end{exx}

\begin{exx}{\hassolution\complex\sref{energypower}
\gref{directproportion}\gref{energyandpower}}
Chilling out by the quarry.

On August 16, 2010 \theGlobe{} described a local business's plan to cool
its corporate facility with water from a nearby quarry rather than
with conventional air conditioning.

\begin{quotation}
[Director of facilities] Dondero estimated that the cooling system,
which eliminates the need 
for any type of refrigerant in the building, saves about \$75,000 a
year, reduces annual water use by one million gallons, and cuts yearly
energy use by about 300,000 kilowatt hours --- enough to power about 30
homes.%
\begin{csmr}
E. Ailworth,
Chilling out by the quarry,
\theGlobe{} (August 16, 2010),
\url{www.boston.com/business/technology/articles/2010/08/16/chilling_out_by_the_quarry/}
\access{March 29, 2020}.
\csmrcomment{Globe, OK}
\end{csmr}
\end{quotation}

\begin{teacher}
The article on the quarry water cooling system also asserts that
\begin{quotation}
The system ... cost only about \$700,000 more than a traditional
cooling system, meaning Biogen Idec should get a return on its
investment in eight to ten years. 
\end{quotation}
Discussing payback time might be interesting --- or too difficult.
\end{teacher}

\begin{abcd*}
\item What rate in dollars per kwh is Dondero using to support his
assertion that this change will save \$75,000 a year?

\item Is the claim that 300,000 kilowatt-hours would power 30 homes
for a year reasonable?
\end{abcd*}

\begin{sol}

\begin{abcd*}
\item What rate in dollars per kwh is Dondero using to support his
assertion that this change will save \$75,000 a year?

This is easy if I just work with the units. I want dollars per kwh so
I divide
\begin{equation*}
\frac{\$75,000}{300,000 \text{ kwh}}
=
0.25
\frac{\$}{\text{kwh}}.
\end{equation*}

I didn't even need a calculator to find that nice round number.

\item Is the claim that 300,000 kilowatt-hours would power 30 homes
for a year reasonable?
\end{abcd*}

There are lots of websites that begin to answer this question. I
wanted one I thought might be authoritative, so I started at the
U.S. Energy Information Administration FAQ at
\url{www.eia.gov/tools/faqs/index.cfm}. There the answer to the
question ``How much electricity does an American home use?'' is

\begin{quotation}
In 2008, the average annual electricity consumption for a
U.S. residential utility customer was 11,040 kwh
\end{quotation}

The newspaper article assumes 10,000 kwh per household per year for 30
homes. That's the right order of magnitude.


\end{sol}
\end{exx}

\begin{exx}{\hassolution\sref{energypower}\gref{directproportion}
\gref{energyandpower}} 
Energy savings at \myindex{MIT}.

On March 26, 2011 Jon Coifman wrote in \theGlobe{} that

\begin{quotation}
In just 36 months, [MIT and NStar] plan to cut the university's energy
use 15 percent --- enough to power 4,500 Massachusetts homes for a
year.%
\begin{csmr}
J. Coifman,
An energy program too efficient for its own good,
\theGlobe{} (March 26, 2011),
\url{www.boston.com/bostonglobe/editorial_opinion/oped/articles/2011/03/26/an_energy_program_too_efficient_for_its_own_good/}
\access{July 23, 2015}.
\csmrcomment{Globe, OK}
\end{csmr}
\end{quotation}

\begin{abcd}

\item If all of MIT's energy use were devoted to powering
Massachusetts homes, how many homes would that be?

\item Compare your answer in part (a) to the number of homes in
Cambridge.

\item 
Estimate MIT's
total annual energy use, in Btus. 

\item Convert your answer to the previous question from Btus to watt-hours.

\end{abcd}

\begin{sol}

\begin{abcd}

\item If all of MIT's energy use were devoted to powering
Massachusetts homes, how many homes would that be?

It follows from the article that MIT uses enough energy in a year
to power $4,500/0.15 = 30,000$ homes. 

\item Compare your answer in part (a) to the number of homes in
Cambridge.

Cambridge has about 100,000 people, so about 30,000 homes. That means
MIT uses as much energy as all the rest of the city (domestically).

Wikipedia says Cambridge has about 40,000 homes, but my estimate is
the same order of magnitude, so the conclusion remains the same.

\item Estimate MIT's
total annual energy use, in Btus. 

Wikipedia says annual household energy use in the
Northeast is about 180,000,000 Btu, half of which is electricity
(\url{en.wikipedia.org/wiki/File:US_household_energy_usage.png}).

\begin{align*}
30,000 &  \text{ households} 
\times
180,000,000 \frac{\text{Btu}}{\text{household}} \\
& \quad = 5,400,000,000,000 {\text{ Btu}} \\
& \quad = 5.4 \text{ teraBtu}.
\end{align*}


\item Convert your answer to the previous question from Btus to watt-hours.
Write your answer using the appropriate metric prefix, not with lots
of zeroes. 

1 kilowatt-hour = 3412.1416 Btu, so
\begin{equation*}
5.4 \times 10^{12} \text{ Btu} \times 
\frac{ 1 \text{ kilowatt-hour}}{3412.1416 \text{ Btu}}
= 1.58 \text{ terawatt-hours}.
\end{equation*}

\end{abcd}

\end{sol}


\end{exx}


\begin{exx}{\hassolution\sref{energypower}\gref{directproportion}
\gref{energyandpower}} 
The Governor gets the units wrong.

On June 7, 2011 \emph{The Norwich Bulletin} reported that

\begin{quotation}
Connecticut Governor Daniel Malloy recently signed off on a
  deal to tax electricity generators one quarter of one cent per
  kilowatt hour, or 25 cents per \$100.%
\begin{csmr}
Attorney General: Connecticut electricity tax could cost Mass. \$26M,
\emph{The Norwich Bulletin} (June 7, 2011),
\url{www.norwichbulletin.com/x832282338/Attorney-General-Connecticut-electricity-tax-could-cost-Mass-26M}
\access{July 23, 2015}.
\csmrcomment{27 words fair use}
\end{csmr}
\end{quotation}

\begin{abcd}

\item What is wrong with the units in this quotation?

\item Estimate the percentage change in the cost of electricity that
would result from a one-quarter of one cent increase per
kilowatt-hour.

\item What do you think Governor Malloy intended to say?

\end{abcd}


\begin{hint}
What did electricity cost in the spring of 2011?
\end{hint}

\begin{sol}

\begin{abcd}

\item What is wrong with the units in this quotation?

The quote says that some number of cents per kilowatt-hour is
equivalent to some number of cents per dollar. That makes no sense at
all.

\item Estimate the percentage change in the cost of electricity that
would result from a one-quarter of one cent increase per
kilowatt-hour.

In the spring of 2011 a kilowatt-hour costs about 10 cents in
Massachusetts.  I need to find out what percentage 0.25 cents (a
quarter of a cent) is of 10 cents. That's easy: $0.25/10 = 0.025 =
2.5\%$.

\item What do you think Governor Malloy intended to say?

I think the Governor meant to say that a quarter of a cent per
kilowatt-hour is 25 cents per hundred kilowatt-hours, not per hundred
dollars.
\end{abcd}

\end{sol}
%
\end{exx}


\begin{movetoextra}{Your total federal tax bill.}
\begin{ExtraExercises}
\begin{exx}{\untested\sref{taxes}\gref{directproportion}
\gref{linearexcel}\gref{incometax}}
Your total federal tax bill.

Modify the graph in Figure~\ref{fig:effectivetax} to show how total
tax and the effective tax rate for (income tax + Social Security)
depends on income.

\end{exx}
\end{ExtraExercises}
\end{movetoextra}


\begin{exx}{\hassolution\complex\sref{taxes}\gref{directproportion}
\gref{incometax}}
President Obama's income tax. 
\index{Obama, Barack}

According to the White House website

\begin{quotation}
[The President] and the First Lady filed their [2013] income tax
  returns jointly and reported adjusted gross income of \$481,098. The
  Obamas paid \$98,169 in total tax. 

The President and First Lady also reported donating \$59,251 --- or about
12.3 percent of their adjusted gross income --- to 32 different
charities. The largest reported gift to charity was \$8,751 to the
Fisher House Foundation. The President's effective federal income tax
rate is 20.4 percent. \ldots  The President and First Lady also
released their Illinois income tax return and reported paying \$23,328
in state income tax.%
\begin{csmr}
J. Carney,
President Obama and Vice President Biden's 2013 Tax Returns,
The White House Blog (April 11, 2014),
\url{www.whitehouse.gov/blog/2014/04/11/president-obama-and-vice-president-biden-s-2013-tax-returns}
\access{July 23, 2015}.
\csmrcomment{government blog no permission needed}
\end{csmr}
\end{quotation}

The President itemized deductions, so he could deduct charitable
contributions and state tax from his adjusted gross income:

\begin{quotation}
In the United States income tax system, adjusted gross income (AGI) is
an individual's total gross income minus specific deductions. Taxable
income is adjusted gross income minus allowances for personal
exemptions and itemized deductions.%
\begin{csmr}
Adjusted gross income, 
Wikipedia,
\url{en.wikipedia.org/wiki/Adjusted_gross_income}
\access{November 14, 2015},
(Creative Commons Attribution-ShareAlike License,
\url{en.wikipedia.org/wiki/Wikipedia:Text_of_Creative_Commons_Attribution-ShareAlike_3.0_Unported_License}).
\end{csmr}
\end{quotation}

\begin{abcd}

\item With the information given, what is the largest possible value
  for the President's taxable income? What tax bracket would he be in?

\item Use the Married Filing Jointly brackets and rates in 
the spreadsheet at 
\link{Federalindividualratehistory.xlsx} 
to compute the Obamas' 2013 federal income tax bill for your answer
to part (a).
If your result
does not match the reported figure, what might explain the difference?

\item If the Obamas had not made those charitable contributions the money
would be part of their taxable income. Use your answers to the
previous questions to answer these.

\begin{abcd*}

\item What would their taxable income have been? What bracket would
  that have put them  in? What would their tax have been?

\item What fraction of the contribution was (essentially) made by the
government?

\item What fraction of the Obamas' income did they contribute to charity? 
\item Did they tithe? 
\item How does their contribution compare to the national average?
\end{abcd*}

\end{abcd}

\begin{sol}

You can find the arithmetic for my solution in the spreadshee
\slink{PresidentsTaxSolution.xlsx}

\begin{abcd}


\item With the information given, what is the largest possible value
  for the President's taxable income? What tax bracket would he be in?

By subtracting the Illinois state tax and charitable contributions
from the adjusted gross income I found a taxable income of 
\$398,519. That's just barely into the 35\% bracket.

\item Use the Married Filing Jointly brackets and rates in 
the spreadsheet at 
\link{Federalindividualratehistory.xlsx} 
to compute the Obamas' 2013 federal income tax bill for your answer
to part (a).
If your result
does not match the reported figure, what might explain the difference?

On that taxable income the tax would be \$107,828. That's about
\$9,000 more than what the Obamas paid. There were probably other
deductions we don't know about.

\item If the Obamas had not made those charitable contributions the money
would be part of their taxable income. Use your answers to the
previous questions to answer these.

\begin{abcd*}
\item What would their taxable income have been? What bracket would
  that have put them  in? What would their tax have been?

Their taxable income would have been \$457,770, putting them in the
top (39.6\%) bracket. They would have paid \$125,846 in tax.

\item What fraction of the contribution was (essentially) made by the
government?

The government would have collected 
\$18,018 more in tax if the Obamas had not donated \$59,251 to
charity. That means the government ``donated'' 
$\$18,018/\$59,251 = 0.30 = 30\%$ of the Obamas' contribution.

\item What fraction of the Obamas' income did they contribute to charity? 

12.3 percent, as the quote says. I checked the arithmetic.

\item Did they tithe? 

Yes --- it's more than ten percent.

\item How does their contribution compare to the national average?

The website
\url{www.philanthropy.com/article/The-Stubborn-2-Giving-Rate/154691}
reported that

\begin{quotation}
Donations from individuals --- almost \$229-billion [in 2012] \ldots
generally hover at around 2 percent of disposable
income, with occasional blips up or down, according to the Giving USA
Foundation, which helps compile the annual philanthropy report.
\end{quotation}
\end{abcd*}

So the Obama's donated about six times as much, as a proportion of
their income.
\end{abcd}

\end{sol}

\end{exx}

\begin{exx}{\hassolution\sref{taxes}\gref{directproportion}\gref{incometax}}
Using the tax table.

Use Table~\ref{table:2019taxtable} to answer the following questions.

\begin{abcd}

\item Compute the tax due in 2019 on a net taxable income of
\$80K. Show your work.

\item Compute the \myindex{effective tax rate} for that income. 

\item Check your answers with those in the spreadsheet \link{GraduatedTax.xlsx}.

\end{abcd}

\begin{sol}

\begin{abcd}

\item The tax on an income of \$80K is
\begin{equation*}
10\% \times \$9700 + 12\% \times (\$39475-\$9700) + 22\% \times
(\$80000-\$39475)
= \$13,458.50	16.82%
\end{equation*}

\item The effective tax rate is $\$13,458.50/\$80000 = 16.82\%$.

\item The answers match those I get by entering
\$80,000 in  cell \cell{G7}.

\end{abcd}

\end{sol}

\end{exx}

\begin{exx}{\hassolution\sref{taxes}\gref{directproportion}\gref{incometax}}
Taxes and inflation.
\index{inflation}

The spreadsheet \link{Federalindividualratehistory.xlsx} contains a
complete history of income tax brackets and rates from the inception
of the income tax in 1913 through 2013.

\begin{abcd}

\item Compute the tax due in 2003 for a single taxpayer with a net
taxable income of \$30K. What is her effective tax rate?

\item Suppose that taxpayer received raises each year that kept up with
inflation. Use an inflation calculator to calculate her net taxable
income in 2013.

\item Use the 2013 tax tables to compute her tax in 2013. What is her
effective tax rate?

\item 
Compare her 2003 tax and effective tax rate with her 2013 tax
and effective tax rate, taking inflation into account. Has her tax
gone up, or down, or stayed the same?

\end{abcd}

\begin{sol}

\begin{abcd}

\item The tax in 2003 on an income of \$30K was
\begin{equation*}
10\% \times \$7000 + 15\% \times (\$28400-\$7000) + 25\% \times
(\$30000-\$28400)
= \$4310.
\end{equation*}

Her effective tax rate was
$\$4310/\$30000 = 0.143666667 \approx 14.46\%$.

\item 
The inflation calculator at
\url{data.bls.gov/cgi-bin/cpicalc.pl} says that \$30,000 in
2003 has the same buying power as \$37,982.12 in 2013

\item The tax in 2013 on an income of \$37,982 is
\begin{equation*}
\begin{split}
10\% \times \$8925 & + 15\% \times (\$36,250-\$8925) \\
                   & + 25\% \times (\$37,982-\$36,250)
= \$5424.
\end{split}
\end{equation*}

Her effective tax rate is 
$\$5424/\$37,982 = 0.142811 \approx 14.28\%$.

%.10 * 7550 + .15*(30650-7550) + .25 * (32869.57-30650)

\item 
Her tax has gone up by \$1114, or 25.8\%. Her income has gone up by
$\$37982/\$30000 = 1.266$, which is a 26.6\% increase. That's
a little bit more than the percentage increase in her taxes, which is
why her effective tax rate decreased by just 0.18 percentage
points. That's not a lot. The IRS did a pretty good job adjusting the
brackets for inflation.

\end{abcd}

\end{sol}

\end{exx}


\begin{exx}{\hassolution\sref{linearfunction}\gref{linearfunctions}
\gref{slopeintercept}}
\myindex{Pandora} growing fast!

On June 10, 2011, CNN Money reported that the internet music site 
\myindex{Pandora} is adding new users at the rate of one per
second. Between February and April the number of users grew from 80 to
90 million.%
\begin{csmr}
B. Rooney,
Pandora raises IPO target to \$200 million,
CNNMoneyTech (June 10, 2011),
\url{money.cnn.com/2011/06/10/technology/pandora_ipo/index.htm}
\access{July 23, 2015}.
\csmrcomment{paraphrase}
\end{csmr}

Is the slope of one user per second correct based on the February and
April numbers of users?

\begin{sol}


Is the slope of one user per second correct based on the February and
April numbers of users?

For a quick and dirty estimate I'll use 100 days for February, March
and April. Then I asked Google and was told
%
\gc*{(10 million miles) per (100 days) =
1.15740741 miles per second}

So the estimate of one user per second is just about right.
\end{sol}

\end{exx}

\begin{exx}{\hassolution\sref{energypower}\gref{energyandpower}}Bicycle power in
  \myindex{Times Square}.

On December 30, 2012 \theGlobe{} reported on an Associated Press story
about six bicycles that would help illuminate the famed falling ball
in Times Square on New Year's Eve. 

\begin{quotation}
Each bike will generate an average of 75 watts an hour. It takes 
50,000 watts to light up the ball's LEDs.%
\begin{csmr}
Sotomayor will help usher out 2013 in NYC,
Associated Press report in \theGlobe{} (December 30, 2013),
%{\def\UrlBreaks{\do\/\do\=\do\-\do\?\do\_\do\3\do\g}
\url{www.bostonglobe.com/news/nation/2013/12/30/justice-sotomayor-lead-times-square-ball-drop/Nac0Vg3INmXYcLcLIYSz4J/story.html}
%}
\access{July 29, 2015}.
\csmrcomment{22 words + paraphrase. fair use}
\csmrcomment{29 words fair use}
\end{csmr}
\end{quotation}

Unfortunately, the Associated Press reporter is quite confused about
the difference between energy and power. The ``generate \ldots 75
watts an hour'' in the quote makes no sense. We 
think what he or she is trying to say is that while someone is
actually pedaling it each bike could power a 75 watt light bulb. All
six bikes together could light up just 450 watts worth of LEDs.

\begin{abcd}
\item How many bikes would have to be pedaled simultaneously to light
  up all the ball's LEDs?

\item 
Since there are only six bikes, people pedaling during the day will
  store the energy they generate in batteries, which will then
  be used to light the ball. Suppose the lights need to be on for two
  minutes while the ball drops at midnight.

How many hours of pedaling will it take to generate (and save) the
electrical energy needed?
\end{abcd}

\begin{sol}

\begin{abcd}
\item How many bikes would have to be pedaled simultaneously to light
  up all the ball's LEDs?

  Since each bike generates 75 watts (when it's being pedaled) and you
  need 50,000 watts to light the ball, you'd need $50,000/75 = 667$
  bikes (rounding up).

\item 
Since there are only six bikes, people pedaling during the day will
  store the energy they generate in batteries, which will then
  be used to light the ball. Suppose the lights need to be on for two
  minutes while the ball drops at midnight.

How many hours of pedaling will it take to generate (and save) the
electrical energy needed?

To light 50,000 watts worth of LEDs for two minutes requires 
100,000 watt-minutes of electricity.
An hour of pedaling on one bikes will generate 75 watt-hours of
electricity. That's $75 \times 60 = 4,500$ 
watt-minutes. You would need $100,000/4,500 = 22.22$ hours of pedaling
to generate the electricity you need. That's about four hours for 
six bikes. 

I'm surprised that the number is so small. 

I can check this answer by using my answer to the first part. There I
figured out that 667 bikes simultaneously would do the job. To keep
the lights on for two minutes I'd need $2 \times 667  = 1334$
bike-minutes of power. That's 
$1334/60 = 22.23$ bike-hours --- the same answer (off in the second
decimal place because I rounded up).

\end{abcd}

\end{sol}
\end{exx}

\begin{exx}{\hassolution\sref{linearfunction}\gref{linearfunctions}\gref{slopeintercept}}
  Flying twice as far. 

A curious traveler asked this question on \myindex{stack exchange}:

\begin{quotation}
A flight from Los Angeles to Albuquerque is about 2 hours
  but is $\approx 670.2$ miles.

A flight from San Jose to Chicago is 4 hours but is $\approx 1859.0$ miles.

Can anyone explain why the travel time from San Jose to Chicago is not
longer and closer to 5.75 hours? 

If the distance increase by 2, shouldn't the time increase by a factor
of 2 as well?%
\begin{csmr}
sifxtreme,
Explain travel times and distances on flight,
Travel Stack Exchange (April 15, 2014),
\url{travel.stackexchange.com/questions/26083/explain-travel-times-and-distances-on-flight}
\access{July 23, 2015},
\ccbysa{}.
\end{csmr}
\end{quotation}

\begin{abcd}

\item Write a linear model for this question. Takeoff and landing will
  take a fixed amount of time. Actual travel in the air will take time
  proportional to the distance traveled. Think about which of the
  variables (time and distance) is the independent variable, and
  identify the slope and intercept with their units.

\item Use the data in the quotation to estimate the two constants in
  your linear model.

\item Compare your answer to those at the link to the quotation.
\end{abcd}

\begin{sol}

\begin{abcd}
\item 
Write a linear model for this question. Takeoff and landing will
  take a fixed amount of time. Actual travel in the air will take time
  proportional to the distance traveled. Think about which of the
  variables (time and distance) is the independent variable, and
  identify the slope and intercept with their units.

In this problem the dependent variable $y$ is the
travel time, in hours. The independent variable $x$ is
the distance flown, in miles. The intercept $b$ is the time in
minutes for takeoff and landing. The slope is $m$. Its units are
minutes per mile. That's the tricky part of this question --- we're
used to thinking of speed in units time/distance, not distance/time.

The linear model is then the familiar $y = mx + b$.

\item Use the data in the quotation to estimate the two constants in
  your linear model.

I know from the data (after a little rounding) that
%
\begin{equation*}
4 = m \times 1860 + b
\end{equation*}
%
and
%
\begin{equation*}
2 = m \times 670 + b .
\end{equation*}

Subtracting tells me that 
\begin{equation*}
2 = (1860 - 670)m = 1190m \approx 1200m
\end{equation*}
%
so $m$ is about (2/1200) hours per mile, or 600 miles per hour. That's
a pretty good approximation for airline cruising speed. Putting $m =
1/600$ into the first equation tells me that 
%
\begin{equation*}
4 = 1860/600 + b \approx 3 + b
\end{equation*}
%
so $b \approx 1$ hour --- half an hour each for takeoff and landing.

\item Compare your answer to those at the link to the quotation.
\end{abcd}

The link says it's reasonable to estimate the cruising speed at about
500 miles per hour and the takeoff and landing time at half an
hour. Those numbers fit the data too.

\end{sol}

\end{exx}

\begin{exx}{\hassolution\sref{linearfunction}\gref{linearfunctions}\gref{slopeintercept}}
  Express lane? 

In \theGlobe{} on November 27, 2015 you could read that

\begin{quotation}
Amid the holiday grocery shopping madness, every
line feels like the wrong one. And yet, some are wronger than
others. Given equally capable cashiers, you are often better off
bypassing the express lane. Research conducted at a large, unnamed,
California grocery store found that while each item adds 3 seconds to
the check-out time, it takes 41 seconds for a person to move through
the line even before their items are added to the tally. Bottom line:
The big time-consumers are not the items, but the small talk and the
paying, says Dan Meyer, who has a doctorate in math education from
Stanford University.%
\begin{csmr}
B. Teitell,
17 holiday blunders (and how to avoid them),
\theGlobe{} November 27, 2015,
\url{www.bostonglobe.com/lifestyle/2015/11/27/the-mistakes-you-make-this-holiday-season-and-how-avoid-them-maybe/cvAPjtOkhi8QoHcZDTig1L/story.html}
\access{March 29, 2020}.
\end{csmr}
\end{quotation}

Suppose you have 10 items in your cart, so you are allowed to use the
express lane. How much longer must the line there be (compared to the
regular lane) to make the wait in the regular lane less? 

You can answer that question with any strategy that makes sense
to you, as long as you explain what you're thinking. If you need a
starting place, one way is to use these steps: 

\begin{abcd}
\item Write the linear equations showing
  how the time it takes a shopper to check out depends on the number
  of items in her cart. What are the slope and intercept, with their
  units? 

\item Suppose shoppers in the express lane buy 6 items (on
  average), while those in the regular lane buy about 20.
Write the linear equations showing your waiting time in each
  line depends on the number of shoppers ahead of you. 

\item Now work on the main question --- which line should you join
  when you have 10 items in your cart? How much longer must the
  express lane line be to make the wait on the regular lane line less?  

\end{abcd}

\begin{hint}
The answer depends on the relative lengths of the lines, not on the
absolute difference in the lengths.
\end{hint}

\begin{sol}

\begin{abcd}
\item Write the linear equations showing
  how the time it takes a   shopper to check out depends on the number
  of items in her cart. What are the slope and intercept, with their
  units? 

Let $T$ be the time (in seconds) and $I$ the number of items in the
cart. Then the equation for checkout time is
%
\begin{equation*}
T = 3I + 41 .
\end{equation*}

The slope is 3 seconds/item and the intercept is 41 seconds.

\item Suppose shoppers in the express line buy 6 items (on
  average), while those in the regular line buy about 20.
Write the linear equations showing your waiting time in each
  line depends on the number of shoppers ahead of you. 

Using my answer to (a) I see that I will have to wait
3*6 + 41 = 59 seconds for each customer ahead of me in the express
line, so
%
\begin{equation*}
W = 59C ,
\end{equation*}
%
where $W$ is the waiting time in seconds and $C$ is the number of
customers ahead of me. The slope is 59 seconds per customer. The
intercept is 0 since if I'm first in line there's no wait.

The equation for the regular line has slope
3*20 + 41 = 101 seconds per customer:
%
\begin{equation*}
W = 101C .
\end{equation*}

\item Now work on the main question --- which line should you join
  when you have 10 items in your cart? How much longer must the
  express line be to make the wait on the regular line less?  

My first thought is that the single seconds in 59 and 101 are a
distraction, and I will approximate them by 60 and 100 seconds. That
means the wait behind 5 customers on the regular line is the same as
the wait behind 3 customers on the express line --- 300 seconds (5
minutes). To decide which line to join I would mentally 
count the customers in groups of 5 in the express
line and groups of 3 in the regular line.

With arithmetic: if the number of customers on the
express line is more than 5/3 = 1.7 times the number of customers on
the regular line, it's better to wait on the regular line. 

I can reach the same answer with algebra. Let $R$ be the humber of
customers on the regular line and $E$ the number on the express
line. Then I want the regular line when
%
\begin{equation*}
61E > 101R
\end{equation*}
which is the same as
\begin{equation*}
E > \frac{101R}{61} = \frac{101}{81}R \approx 1.7.
\end{equation*}
%\end{hint}
\end{abcd}

\end{sol}

\end{exx}


\begin{ReviewExercises}

\begin{rexx}
If you drive at a rate of 50 miles per hour for 3 hours, how far have
you driven?  Identify each piece of this proportion:  the quantities
being measured and the proportionality constant, with the appropriate
units.  
\end{rexx}

\begin{rexx}You may remember from geometry that the circumference of a
circle is directly proportional to the diameter of that circle.  The
relationship is 
%
\begin{equation*}
c = \pi d, 
\end{equation*}
where $c$ represents the circumference and $d$ represents
the diameter and $\pi \approx 3.14$ is the proportionality constant.
If the diameter of a circle is doubled, how does the circumference
change? 
\end{rexx}

\begin{rexx}
The cost of potatoes is proportional to the weight (in pounds) you buy.
If potatoes cost \$0.69 per pound, what is the cost for 3
pounds of potatoes?  
\end{rexx}

\begin{rexx} 
The conversion from {\textsterling} to U.S. \$ is 1.53
\$/{\textsterling}. How much is {\textsterling}200
worth in U.S.\$? 
\end{rexx}

\begin{rexx}
Suppose that $y$ is directly proportional to $x$.  When $x=16$, then
$y=4$. 
\begin{abcd}
\item What is the proportionality constant?

\item  If $x=32$, what is $y$? 

\item If $y=32$, what is $x$?
\end{abcd}
\end{rexx}

\begin{rexx}
Identify the slope and intercept in each of the following.  When
appropriate, state the units. 
\begin{abcd*}
%CHANGE added periods to items
\item $y = 2.5x + 6$.
\item $y = -5x +20$.
\item $y = 300 + 40x$.
\item $Q = 0.004E - 300$.
\item  He earns \$9.25 per hour.
\item  To rent a car for one day, the cost is \$25 plus \$0.15 per mile.
\item  My new phone cost \$25, plus a monthly charge of \$15.
\item  The conversion from {\textsterling} to U.S. \$ is 1.53
\$/{\textsterling}.
\item  The salesperson worked only on commission,
earning 20\% of the total amount sold. 
\end{abcd*}
\end{rexx}

\begin{rexx}
Solve each problem.
\begin{abcd*}  
\item If $y = 2.5x + 6$ and $x=4$, what is $y$?  
\item  If $y = -5x +20$ and $y=0$, what is $x$?
\item  If $y = 300 + 40x$ and $x=-10$, what is $y$?
\item If his salary is \$9.25 per hour and he works 5 hours, how much
does he earn? 
\item  If the conversion from U.S. dollars to pounds sterling is 1.80
\$/{\textsterling}, how much money would you get by changing \$100 to
{\textsterling}?   
\item  If my new phone cost \$25, and I pay a monthly charge of \$15,
what is my total cost after 10 months?  When does my total 
cost reach \$250?
\end{abcd*}
\end{rexx}

%\begin{rexx} 
%Verify that there are about 9000 hours in one year.  Use that to
%estimate the number of hours in a decade.  Approximately how many
%hours are in a century? 
%\end{rexx}
%

\end{ReviewExercises}

\begin{NewExercises}

  \begin{exx}{\untested\hassolution\sref{taxes}\gref{linearfunctions}\gref{incometax}}
    Instructions from the IRS.

    Figure~\ref{fig:form1040-2018tax} is the form the Internal Revenue
    Service provides for computing 2018 tax on incomes over \$100,000.

    \begin{abcd}

    \item
      Calculate the tax on an income of \$150,000 using this
      form. (That income is the amount on line 10.)

    \item
      Calculate the tax on an income of \$150,000 using the procedure
      in Figure~\ref{fig:2018taxbrackets}. Check that you get the same
      answer.

    \item Explain how the IRS arrived at the figure \$5,710.50 in
      column (d) in the first row of the form.
\end{abcd}
\figfile{form1040-2018tax.pdf}
\begin{figure}
\centering
\includegraphics[width=\textwidth]{\thefigurefilename}
\begin{csmr}[Tax calculation (2018)\label{fig:form1040-2018tax}]
Internal Revenue Service Form 1040 (2018),
p. 79,
\url{www.irs.gov/pub/irs-pdf/i1040gi.pdf} 
\access{February 15, 2019}.
\end{csmr}
\end{figure}
\figfile{}

\begin{sol}

  \begin{abcd}
    \item
      Calculate the tax on an income of \$150,000 using this
      form. (That income is the amount on line 10.)

      The Google calculator tells me

      \gc{150,000 * 0.24 - 5,710.50 = 30289.5}

      so the tax is \$30,289.50.
      
    \item
      Calculate the tax on an income of \$150,000 using the procedure
      in Figure~\ref{fig:2018taxbrackets}. Check that you get the same
      answer.

      The Google calculator tells me

      \gc{(150,000 - 82,500) * 0.24 + 14,089.50 = 30289.5}

      which is the same answer as in (a).
      
    \item Explain how the IRS arrived at the figure \$5,710.50 in
      column (d) in the first row of the form.

    The instructions ask you to calculate the tax as if all the income
    was at your highest bracket, and then subtract an amount that
    corrects for the fact that some of your income should be taxed at
    a lower rate.
\end{abcd}

\end{sol}
\end{exx}
\end{NewExercises}

\setexercisecounter{}

\begin{ExtraExercises}



\begin{exx}{\untested\complex\needsquestions} 
Does virtual save energy?

\begin{quotation}
MOST PEOPLE take for granted the Earth-friendly nature of
  electronic communication. Paperless, ink-free, no shipping supplies,
  no gas for transportation: the environmental benefits of virtual
  communication are obvious. But the reality is more complicated, at
  least according to a growing number of concerned technology experts
  and scientists. Vast stockpiles of digital data waste energy, too. 

Everyday emails aren't to blame. But large photo and video
attachments, cluttered inboxes, and massive email forwards may
be. Some analysts estimate that emailing a 4.7-megabyte attachment ---
the equivalent of four large digital photos --- can use as much energy
as it takes to boil about 17 kettles of water. The problem is
magnified when large emails are forwarded to many people and left in
inboxes undeleted. As long as emails remain in your inbox, the data
they create is physically stored somewhere. 

And that's where the problems arise: The total amount of digital
storage worldwide is approaching 1 zettabyte, or 1 million times the
contents of the Earth's largest library. Currently, that information
is archived on equipment with a mass equivalent to 20 percent of
Manhattan. Global data storage is expected to reach 35 zettabytes by
2020, which means more equipment, land, and energy. The information
industry already accounts for approximately 2 percent of global carbon
dioxide emissions. That's the same amount as the airline industry
blasts into the atmosphere. Coupled with the rapid increase in stored
data, it's an unsustainable scenario. 

Technology firms must create systems that store data with less energy,
and governments should provide incentives for them to do so. Just as
important, consumers must demand products that save energy, and use
websites like Flickr and MediaFire that allow them to share large
files without emailing. Better still, they could consider keeping some
of those embarrassing photos and home videos to themselves.%
\begin{csmr}
Don't forward those photos
\theGlobe{} (September 7, 2010),
\url{www.boston.com/bostonglobe/editorial_opinion/editorials/articles/2010/09/07/dont_forward_those_photos/}
\access{July 23, 2015}.
\csmrcomment{This is a really long quote with lots of numbers, no
  questions yet. Since we have blanket permission from the Globe for
  this kind of material, we can quote it here.}
\end{csmr}
\end{quotation}

\end{exx}

\begin{exx}{\untested\needsquestions}
Green gas?

Robert Bryce's op-ed in \theTimes{} on
June 7, 2011
(\url{www.nytimes.com/2011/06/08/opinion/08bryce.html})
has lots of interesting numbers about the costs in steel and land area
for solar and wind electricity generation.

\end{exx}

\begin{exx}{\untested\needsquestions} 
Every little bit counts.

On March 5, 2012 \theGlobe{} reported on the
Ocean Renewable Power Company's plans to install tidal powered
generators in Maine:
\index{tidal power}

\begin{quotation}
The first unit capable of powering 20 to 25 homes will be
hooked up to 
the grid this summer, and four more units will be installed next year
at a total cost of \$21 million \ldots

Eventually, Ocean Renewable hopes to install more units to bring its
electrical output to 4 megawatts.%
\begin{csmr}
D. Sharp,
Maine company ready to install tidal power unit,
Associated Press reported in \theGlobe{}  (March 05, 2012),
\url{www.bostonglobe.com/business/2012/03/05/maine-company-ready-install-tidal-power-unit/daJ3ivfrUNUnHejoOt2lqJ/story.html}
\access{July 23, 2015}.
\csmrcomment{Globe so OK}
\end{csmr}
\end{quotation}

\end{exx}


\begin{exx}{\hassolution\artificial\sref{comparelinear}\gref{linearfunctions}\gref{slopeintercept}} 
Compact fluorescent bulbs.

Consumers are being encouraged to replace ordinary light bulbs
with compact fluorescent bulbs.\index{compact fluorescent bulb}
(CFLs). Soon they will be required to.

(This exercise should be updated to discuss LED bulbs too, and to use
real rather than invented numbers.)

A CFL uses less energy than an ordinary
incandescent bulb that produces the same amount of light, but it costs
more to buy. This table provides data with which you can
compare the two. 

\begin{center}
\begin{tabular}{ccc}
\toprule
bulb & initial cost & power \\
\midrule
ordinary & \$2.00 & 100 watts \\
CFL & \$9.00 & 25 watts \\
\bottomrule
\end{tabular}
\end{center}

Suppose electricity costs \$0.20 per kwh.

You can use pencil and paper, a calculator or Excel to do the
arithmetic.

\begin{abcd}

\item Write a linear equation with which you can calculate the total
cost $C$ of using the ordinary bulb for $H$ hours.

\item What is the slope of that equation (with its units)?

\item What is the intercept of that equation (with its units)?

\item How much would it cost to buy the ordinary bulb and use it for
1000 hours?

\item Write a linear equation with which you can calculate the total
cost $C$ of using the CFL for $H$ hours.

\item How much would it cost to buy the CFL and use it for
1000 hours?

\item How long would you have to use the CFL to make it worth having
paid the higher purchase price?
\end{abcd}

\item Are the five numbers given in this exercise reasonable?

\item What does ``incandescent'' mean? Why are incandescent light
bulbs called that?

\begin{sol}


\begin{abcd}
\item Write a linear equation with which you can calculate the total
cost $C$ of using the ordinary bulb for $H$ hours.

\begin{equation*}
C = 2.00 + 0.1\times 0.20 H .
\end{equation*}
\item What is the slope of that equation (with its units)?

The slope is 0.02 \$/hour, or 2 cents/hour.

\item What is the intercept of that equation (with its units)?

The intercept is \$2.00.

\item How much would it cost to buy the ordinary bulb and use it for
1000 hours?
\begin{equation*}
C = \$2.00 + 0.02 \frac{\$}{\text{hour}}\times 1000 \text{ hours}
= \$22.00.
\end{equation*}
\item Write a linear equation with which you can calculate the total
cost $C$ of using the CFL for $H$ hours.

\begin{equation*}
C = 9.00 + 0.025\times 0.20 H .
\end{equation*}

\item How much would it cost to buy the CFL and use it for
1000 hours?

\begin{equation*}
C = \$9.00 + 0.005 \frac{\$}{\text{hour}}\times 1000 \text{ hours	}
= \$14.00.
\end{equation*}

\item How long would you have to use the CFL to make it worth having
paid the higher purchase price?

From my previous work, the answer will be less than 1000 hours. I can
find the exact time by algebra or by trial and error. It turns out to
be 467 or about 500 hours. That makes sense --- the CFL costs a penny
and a half less per hour to run, which means it will take about 500
hours to save the \$7 difference in initial cost.

\item Are the five numbers given in this exercise reasonable?

Waiting for student input.

\item What does ``incandescent'' mean? Why are incandescent light
bulbs called that?

``Incandescent'' means 
``giving off light because it's hot''. That's just how ordinary old
fashioned light bulbs work. There's a thin filament (wire) inside that
heats up and glows.

\end{abcd}

\end{sol}

\end{exx}


\begin{exx}{\untested\hassolution\sref{energypower}\gref{energyandpower}}
Not flying to London.

In \theTimes{} on April 25, 2011 you could read that

\begin{quotation}
You can save so much energy by not flying to London [to collaborate
  with a coworker] that it will run a rack of computers for a year.%
\begin{csmr}
J. Markoff,
Digging Deeper, Seeing Farther: Supercomputers Alter Science,
\theTimes{} (April 25, 2011),
\url{www.nytimes.com/2011/04/26/science/26planetarium.html}
\access{July 23, 2015}.
\csmrcomment{22 words fair use}
\end{csmr}
\end{quotation}

Estimate the energy costs of flying to London and running a rack of
computers for a year to see if they are of the same order of
magnitude.

If you're a physicist you can make these estimates with your common
knowledge. If you're not, you can put together reliable information
from the web. Try searching for the energy cost of flying an airplane
and the energy cost of running a computer.

\begin{sol}
From \url{www.inference.phy.cam.ac.uk/withouthotair/c5/page_35.shtml}:

\begin{quotation}
Imagine that you make one intercontinental trip per year by plane. How
much energy does that cost?

A Boeing 747-400 with 240,000 litres of fuel carries 416 passengers
about 8,800 miles (14,000 km). And fuel's calorific value is 10 kwh
per litre. (We learned that in Chapter 3.) So the energy cost of one
full-distance roundtrip on such a plane, if divided equally among the
passengers, is 
\begin{equation*}
\frac{2 \times 240,000 \text{ litre}}{416 \text{ passengers}}
 	\times 10 \text{ kwh/litre} \approx  12,000 \text{ kwh per passenger}.
\end{equation*}
If you make one such trip per year, then your average energy consumption
per day is 
\begin{equation*}
\frac{12,000 \text{ kwh}}{365 {\text{ days}}}
\approx  33 \text{ kwh/day}.
\end{equation*}
\end{quotation}

The round trip airline distance from Boston to London is about 6,500
miles, so that trip will cost somewhat less than the trip above, 
figure 24 kwh/day. 

At (say) 100 watts to run a computer (that's the right order of
magnitude) you use about 2,400 watt-hours or 2.4 kwh in a day. So 24
kwh will power ten computers for a day. That's a pretty small
rack, but the order of magnitude is right.

\end{sol}
\end{exx}

\begin{exx}{\hassolution}
Express lane?

In \theGlobe{} on November 27, 2015 you could read that

\begin{quotation}
Amid the holiday grocery shopping madness, every
line feels like the wrong one. And yet, some are wronger than
others. Given equally capable cashiers, you are often better off
bypassing the express lane. Research conducted at a large, unnamed,
California grocery store found that while each item adds 3 seconds to
the check-out time, it takes 41 seconds for a person to move through
the line even before their items are added to the tally. Bottom line:
The big time-consumers are not the items, but the small talk and the
paying, says Dan Meyer, who has a doctorate in math education from
Stanford University. 
\begin{csmr}
B. Teitel, 
17 holiday blunders (and how to avoid them),
\theGlobe{}, November 27, 2015,
\url{www.bostonglobe.com/lifestyle/2015/11/27/the-mistakes-you-make-this-holiday-season-and-how-avoid-them-maybe/cvAPjtOkhi8QoHcZDTig1L/story.html}
\access{November 30, 2015}.
\end{csmr}
\end{quotation}

Suppose you have 10 items in your cart, so you are allowed to use the
express lane. How much longer must the line there be (compared to the
regular lane) to make the wait in the regular lane less? 

Hint:
The answer depends one the relative lengths of the lines, not on the
absolute difference in the lengths.

You can answer that question with any strategy that makes sense
to you, as long as you explain what you're thinking. If you need a
starting place, one way is to follow these steps: 

\begin{abcd}
\item Write the linear equation showing
  how the time it takes a shopper to check out depends on the number
  of items in her cart. What are the slope and intercept, with their
  units? 

\item Suppose shoppers in the express lane buy 6 items (on
  average), while those in the regular lane buy about 20.
Write a linear equation for each line  showing your waiting time
depends on the number of shoppers ahead of you.  


\item Now work on the main question --- which line should you join
  when you have 10 items in your cart? How much longer must the
  express lane line be to make the wait on the regular lane line less?  

\end{abcd}

\begin{sol}


\begin{abcd}
\item Write the linear equations showing
  how the time it takes a   shopper to check out depends on the number
  of items in her cart. What are the slope and intercept, with their
  units? 

Let $T$ be the time (in seconds) and $I$ the number of items in the
cart. Then the equation for checkout time is
%
\begin{equation*}
T = 3I + 41 .
\end{equation*}

The slope is 3 seconds/item and the intercept is 41 seconds.

\item Suppose shoppers in the express line buy 6 items (on
  average), while those in the regular line buy about 20.
Write the linear equations showing your waiting time in each
  line depends on the number of shoppers ahead of you. 

Using my answer to (a) I see that I will have to wait
3*6 + 41 = 59 seconds for each customer ahead of me in the express
line, so
%
\begin{equation*}
W = 59C ,
\end{equation*}
%
where $W$ is the waiting time in seconds and $C$ is the number of
customers ahead of me. The slope is 59 seconds per customer. The
intercept is 0 since if I'm first in line there's no wait.

The equation for the regular line has slope
3*20 + 41 = 101 seconds per customer:
%
\begin{equation*}
W = 101C .
\end{equation*}

\item Now work on the main question --- which line should you join
  when you have 10 items in your cart? How much longer must the
  express line be to make the wait on the regular line less?  

My first thought is that the single seconds in 59 and 101 are a
distraction, and I will approximate them by 60 and 100 seconds. That
means the wait behind 5 customers on the regular line is the same as
the wait behind 3 customers on the express line --- 300 seconds (5
minutes). To decide which line to join I would mentally 
count the customers in groups of 5 in the express
line and groups of 3 in the regular line.

With arithmetic: if the number of customers on the
express line is more than 5/3 = 1.7 times the number of customers on
the regular line, it's better to wait on the regular line. 

I can reach the same answer with algebra. Let $R$ be the humber of
customers on the regular line and $E$ the number on the express
line. Then I want the regular line when
%
\begin{equation*}
61E > 101R
\end{equation*}
which is the same as
\begin{equation*}
E > \frac{101R}{61} = \frac{101}{81}R \approx 1.7 R.
\end{equation*}
%\end{hint}
\end{abcd}

\end{sol}

\end{exx}

\begin{exx}{\untested}
LED lightbulb.

Figure~\ref{fig:ledbulb} shows the specifications for an LED bulb that
costs \$11.00. It's meant to replace a 60 watt incandescent bulb that
costs about a dollar.  

\figfile{ledbulb.pdf}
\begin{figure}
\centering
\framebox{
  \includegraphics[width=4in]{\thefigurefilename}
}
\caption{Light bulb specs}
\label{fig:ledbulb}
\end{figure}
\figfile{}

\begin{abcd}
\item Check the claimed Estimated Yearly Energy Cost.

\item Check that the Rated Life agrees with the claimed 22.8 year
  Life.

\item How long will it take for the LED bulb to be cheaper overall
  than the incandescent bulb
  (counting both the initial cost and the cost of the electricity to
  run it) ?

\item What does the acronym ``LPW'' stand for? Has it been computed
  correctly?

\item What do the words ``LED'' and ``incandescent'' mean?

\end{abcd}

\end{exx}

\begin{exx}{\hassolution}
Kilowatt hours per day.

Professor Sir David Mackay's
\index{MacKay, David}
obituary in \emph{The Telegraph} observed that

\begin{quotation}
  [His] genius was to express all
forms of power consumption and production in a single unit of
measurement --- kilowatt hours per day (kWh/d). A 40 watt lightbulb,
kept switched on all the time, uses one kWh/d, while driving the
average car 50km a day consumes 40 kWh/d.  
\begin{csmr}
Professor Sir David MacKay, physicist --- obituary,
\emph{The Telegraph}, April 15, 2016,
\url{www.telegraph.co.uk/obituaries/2016/04/15/professor-sir-david-mackay-physicist--obituary/}
\access{May 1, 2016}.
\end{csmr}
\end{quotation}

\begin{abcd}

\item Confirm that keeping a 40 watt lightbulb left switched on all
  day would use 1 kilowatt hour of energy.  

\item According to MacKay, ``The amount of energy saved by switching off
  the phone charger [for a day] is exactly the same as the energy used
  by driving an average car for one second.'' How many people would
  have to switch of their cell phone chargers in a day to save as much
  energy as driving a car 50 km? 

\item Mackay claims that switching your cell phone charger off for
  one year saves as much  energy as is needed for one hot bath. How 
  many kilowatt hours per day does it take to heat that bath water?
  water? 

\item What did MacKay mean when he stated that gestures like turning
  of your cell phone charger were akin to ``bailing out the Titanic
  with a teaspoon''? Do you agree? 

\end{abcd}

\begin{sol}
\begin{abcd}

\item Confirm that keeping a 40 watt lightbulb left switched on all
  day would use 1 kilowatt hour of energy.  

\begin{equation*}
40 \text{ watts} \times 24 \text{ hours} = 960 \text{ watt-hours}
\approx 1 \text{ kilowatt-hour}.
\end{equation*}

\item According to MacKay, ``The amount of energy saved by switching off
  the phone charger [for a day] is exactly the same as the energy used
  by driving an average car for one second.'' How many people would
  have to switch of their cell phone chargers in a day to save as much
  energy as driving a car 50 km? 

I know from the quote that it takes one kilowatt-hour to drive a car 50
km. To figure out how much energy it takes to drive it for one second
I need to estimate the speed of the car. To make the arithmetic easy
I'll assume it's driving 50 km/hour (about 30 miles/hour). So the car
needs 1 kilowatt-hour to drive for an hour. 

MacKay says that each phone charger's energy savings could drive the
car for a second, so it would take turning about 3,600 of them off for
a day to save the energy to drive the car 50 km.

\item MacKay claims that switching your cell phone charger off for
  one year saves as much  energy as is needed for one hot bath. How 
  many kilowatt-hours of energy does it take to heat that bath water?

Switching off the cell phone charger for a day saves 1/3,600 of a
kilowatt-hour. Multiplying by (approximately) 360 days per year leads
to a savings of 0.1 kilowatt-hours. So that's how much energy it takes
to heat the bath water.

\item What did MacKay mean when he stated that gestures like turning
  of your cell phone charger were akin to ``bailing out the Titanic
  with a teaspoon''? Do you agree? 

He means that the gesture is just a gesture, and won't make any
difference when it comes to solving the real problem --- saving the
titanic or conserving energy. I agree with the conclusion. I do wonder
if he thought about this quantitatively --- how many people with
teaspoons would it take to bail out the Titanic?
\end{abcd}

\end{sol}

\end{exx}

\begin{exx}{\needsquestions}
The Rosenfeld.
\index{Rosenfeld, Robert}

From \theTimes{} January 27 obituary of physicist Arthur Rosenfeld:
\begin{quotation}
[In 2010] a group of scientists proposed a unit of measurement
in his name. The ``Rosenfeld,'' they said, should refer to annual
electricity savings of three billion kilowatt-hours --- enough to
eliminate the need for a coal plant.
\begin{csmr}
K. Galbraith,
Arthur Rosenfeld, Zealous Champion of Energy Efficiency, Dies at 90,
\emph{The New York Times}, January 27, 2017,
\url{www.nytimes.com/2017/01/27/science/arthur-rosenfeld-dead-energy-efficiency.html}
\access{February 8, 2017}.
\end{csmr}
\end{quotation}

\end{exx}

\begin{exx}{\untested\needsquestions}
Fuel economy in square meters?

\url{physics.stackexchange.com/questions/325733/why-can-fuel-economy-be-measured-in-square-meters}

\begin{quotation}
The fact that the gallon (or liter) is a volume is really
irrelevant. What we're really measuring is distance per energy
unit. It's just handy that all kinds of gasoline have pretty close to
the same energy content for a given volume (about 32.8 kWh/gal, per
Google), and technology/commerce makes it convenient to measure \&{} sell
liquids by volume. (But other liquid fuels, say ethanol or diesel,
have kWh/volume, so we get different mpg from them.) So plug in the
kWh/gal figure into your math, and get miles/kWh :-)  
\end{quotation}

And of course

\url{what-if.xkcd.com/11/}
\index{xkcd}

\end{exx}

\begin{exx}{\needsquestions\worthy}
Home solar power.
\index{solar power}
From \theGlobe{} on May 28, 2017:

\begin{quotation}
The median price per watt of capacity for systems financed through the
Mass Solar Loan Program is \$4, and the median installation size is 8.1
kilowatts. Those numbers suggest a new solar system could easily run
more than \$30,000. 

Fortunately, several programs can help defray the costs. Federal tax
incentives allow you to take a credit worth 30 percent of the cost of
the installation. Massachusetts also offers a tax credit of 15 percent
of the remaining cost after the federal incentive has been subtracted,
with a maximum value of \$1,000. 

Together, these credits could reduce the cost of a \$30,000 system to
roughly \$20,000 --- still a pretty hefty total.%
\begin{csmr}
S. Shemkus,
Home solar power becomes a brighter prospect for many,
\theGlobe{}, May 28, 2017,
\url{www.bostonglobe.com/business/2017/05/28/shedding-light-home-solar-systems/2eZClXIj20e36yha8v2dwN/story.html}
\access{June 1, 2017}.
\end{csmr}

\end{quotation}
\end{exx}

\begin{exx}{\untested\needsquestions} 
What does a butcoin cost?
\index{bitcoin}

On January 22, 2018 you could read in \theTimes{} that

\begin{quotation}
\ldots the computer power needed to create each digital token consumes at
least as much electricity as the average American household burns
through in two years \ldots%
\begin{csmr}
N. Popperjan,
There Is Nothing Virtual About Bitcoin's Energy Appetite,
\emph{The New York Times},
January 21, 2018,
\url{www.nytimes.com/2018/01/21/technology/bitcoin-mining-energy-consumption.html}
\access{January 22, 2018}
\end{csmr}
\end{quotation}

There are more question opportunities later in the article.
\end{exx}

\begin{exx}{\untested\needsquestions}
What is fair?
\index{fair}
\index{Zuckerberg, Mark}
\index{parking ticket}

In March 2018  
\theTimes{} published an opinion piece headlined
``A Billionaire and a Nurse Shouldn't Pay the Same Fine for Speeding''

\begin{quotation}
If Mark Zuckerberg and a janitor who works at Facebook's headquarters
each received a speeding ticket while driving home from work, they'd
each owe the government the same amount of money. Mr. Zuckerberg
wouldn't bat an eye. 

The janitor is another story.

For people living on the economic margins, even minor offenses can
impose crushing financial obligations, trapping them in a cycle of
debt and incarceration for nonpayment. In Ferguson, Mo., for example,
a single \$151 parking violation sent a black woman struggling with
homelessness into a seven-year odyssey of court appearances, arrest
warrants and jail time connected to her inability to pay. %
\begin{csmr}
A. Schierenbeck
A Billionaire and a Nurse Shouldn't Pay the Same Fine for Speeding,
\emph{The New York Times}, March  15, 2018,
\url{www.nytimes.com/2018/03/15/opinion/flat-fines-wealthy-poor.html}
\access{March 17, 2018}.
\end{csmr}

Read the article and the reader's comments. If you were to post a
comment at the website what would you say? You might consider
comparing the fine structure to the graduated income tax.
\end{quotation}
\end{exx}


\begin{exx}{\untested\needsquestions}
Capturing fog.\index{fog}

On June 7, 2018 \theGlobe{} reported on an MIT project to capture the
water in fog.

\begin{quotation}
Varanasi said a typical 600-megawatt power plant consumes as much
water annually as a city of 100,000 people, losing 750 million gallons
to the air, 20 to 30 percent of which would be fog droplets. 

Varanasi envisions his system capturing 150 million gallons from the
hypothetical plant. Varanasi didn't go into the details of what the
system installation and operation would cost but said it would need
just \$10,000 in electricity to run.%
\begin{csmr}
M. Finucane,
MIT says new system captures water in fog,
\theGlobe, June 8, 2018,
\url{www.bostonglobe.com/metro/2018/06/08/mit-says-new-system-could-someday-allow-people-have-nice-cold-glass-fog/bwECnMe9OYEF7pwfIKhXqI/story.html}
\access{July 30, 2019}
\end{csmr}
\end{quotation}

\begin{abcd}
\item Estimate or search to find out how much water 100,000 people use
  annually. 

At their website \url{water.usgs.gov/edu/qa-home-percapita.html} the
  USGS says that  ``Estimates vary, but each person uses about 80-100
  gallons of water   per day.''

  Calculating with 350 days per year 100,000 people use between
  2.8 and 3.5 billion gallons per year.
    
\item What fraction of the water consumed by the power plant is lost?

 750 million is one fourth of 3.5 billion, so about 25 percent of the
 water is lost.

 I wonder where the other $3/4$ goes.
 
\item What fraction of the lost water would this system capture?

  150 million is one fifth of 750 million.
  
\item What is the cost in electricity to capture one gallon of the
  water?

  \begin{equation*}
    \frac{\$10,000}{150 \times 10^6 \text{ gallons}}
         =
         0.00007 \frac{\$}{\text{gallon}}
\end{equation*}
\end{abcd}
\end{exx}
\end{ExtraExercises}



