\chapter{Functions and graphs}
\section{Applications}
\subsection{Geometric relationships as functions}
Many formulas from Geometry, Business, Physics and other applications are nothing but functions
that define one quantity when we are given one (or more) other quantities. For example, the formula
\[C=2\pi r\]
for the circumference of a circle is a function with independent variable \(r\)
(the radius of the circle)
and dependent variable \(C\) (the circumference). In this case, we say
that the circumference is a \textit{function of} the
radius. In the same way, the formula,
\[A=\pi r^2\]
expresses the area \(A\) of a circle as a function of its radius \(r\), and the formula,
\[P=4x\]
expresses the perimeter \(P\) of a square as a function of its side \(x\).
Listed below are some of the formulas you should know to be able to solve application problems:
Area
\(
\begin{array}{|c|c|c|c|}
\hline
\mbox{Shape} & \mbox{Rectangle} & \mbox{Triangle} & \mbox{Circle}\\
\hline
\mbox{Formula} & A=LW & \displaystyle A=\frac{1}{2}bh & A=\pi r^2\\[1ex]
\hline
\end{array}
\)
\(L=\) Length, \(W=\) Width, \(b=\) base, \(h=\) height, \(r=\) radius
Perimeter
\(
\begin{array}{|c|c|c|c|}
\hline
\mbox{Shape} & \mbox{Rectangle} & \mbox{Triangle} & \mbox{Circle}\\
\hline
\mbox{Formula} & P=2L+2W & P=a+b+c & C=2\pi r\\[1ex]
\hline
\end{array}
\)
\(a,b,c=\) sides of triangle
Volume
\(
\begin{array}{|c|c|c|}
\hline
\mbox{Shape} & \mbox{Box} & \mbox{Cylinder} \\
\hline
\mbox{Formula} & V=LWH & V=\pi r^2h\\[1ex]
\hline
\end{array}
\)
\(L=\) Length, \(W=\) Width, \(H=\) Height
Straight lines
\(
\begin{array}{|c|c|c|}
\hline
\mbox{Slope} & \mbox{Slope-Intercept} & \mbox{Point-Slope} \\
\hline
\displaystyle m=\frac{y_2-y_1}{x_2-x_1} & y=mx+b & y-y_1=m(x-x_1)\\
\hline
\end{array}
\)
\((x_1,y_1), (x_2,y_2)\) points on the line, \(m=\) slope, \((0,b)=y\)-intercept.
The following examples show how to derive function relationships between two geometric quantities.
\begin{example}
We express the radius of a circle as a function of its area \(A\). We start from the formula,
\[A=\pi r^2\]
expressing \(A\) as a function of \(r\), and then we solve the equation for \(r\):
\[
\begin{array}{rcll}
A & = & \pi r^2 & \mbox{The formula for the area of a circle}\\
\ \\
\displaystyle{\frac{A}{\color{red}{\pi}}} & = & \displaystyle{\frac{\pi r^2}{\color{red}{\pi}}} &
\mbox{Divide both sides by \(\pi\)}\\
\ \\
\displaystyle{\frac{A}{\pi}} & = & r^2 & \mbox{Simplify}\\
\ \\
\displaystyle{\sqrt{\frac{A}{\pi}}} & = & \sqrt{r^2} & \mbox{Take the square root of both sides}\\
\ \\
r & = & \displaystyle{\sqrt{\frac{A}{\pi}}} & \mbox{Simplify, and switch sides}
\end{array}
\]
The formula on the last line gives us what we want: it expresses the radius \(r\)
as a function of the area \(A\).
Note that we did not use the \(\pm\) sign in front of the square root sign, because we know that the radius \(r\)
must be positive.
\end{example}
\begin{example}
Now we express the area \(A\) of a circle as a function of its
circumference \(C\). First we solve the formula
\(C=2\pi r\) for \(r\):
\[
\begin{array}{rcll}
C & = &2\pi r & \mbox{The formula for the circumference}\\
\ \\
\displaystyle{\frac{C}{\color{red}{2\pi}}} & = & \displaystyle{\frac{2\pi r}{\color{red}{2\pi}}} &
\mbox{Divide both sides by \(2\pi\)}\\
\ \\
r & = & \displaystyle{\frac{C}{2\pi}} & \mbox{Simplify, and switch sides}
\end{array}
\]
Now we substitute what we found in the formula \(A=\pi r^2\):
\[
\begin{array}{rcll}
A & = & \displaystyle{\pi \left(\frac{C}{2\pi}\right)^2}\\
\ \\
& = & \displaystyle{\pi \frac{C^2}{4\pi^2}}\\
\ \\
& = &\displaystyle{ \frac{C^2}{4\pi}}.
\end{array}
\]
So the formula
\[A=\displaystyle \frac{C^2}{4\pi}\]
expresses \(A\) as a function of \(C\).
\end{example}
\subsection{Business applications}
We now discuss some business applications. Remember that the revenue of a business is the total
amount of money that the business brings in, and to find the actual profit we need to subtract the cost
incurred to run the business. We are going to denote by \(C\), \(R\) and \(P\) the Cost, Revenue
and Profit. So the equation that relates them is
\[P=R-C.\]
Often the total cost is made of two parts: the fixed cost (for example the cost of setting up a business), and the variable
cost, that depends on what the business produces. So for example if the business is a bakery that produces loaves of bread,
the variable cost would depend on how many loaves of bread are produced.
\begin{example}
A company produces a product for which the variable cost is \$12 per unit and the fixed costs are \$100,000.
The product sells for \$14. Let \(x\) be the number of units produced and sold. Then the total cost will be a function of \(x\),
and we denote it by \(C(x)\). Since each unit costs \$12 to produce, producing \(x\) units will cost \(12x\). Adding the fixed
cost, we get that the formula for the total cost is
\[C(x) = 12x + 100,\!000.\]
The revenue \(R(x)\) is also a function of \(x\) and it is simple to find: since each unit is sold for \$14, the revenue from
selling \(x\) units will be\[R(x)=14x.\]
Once we have Cost and Revenue, the Profit is just the difference:
\begin{eqnarray*}
P(x)&=&R(x)-C(x)\\
&=&14x-(12x+100,\!000)\\
&=&14x-12x-100,\!000\\
&=&2x-100,\!000
\end{eqnarray*}
Now that we have all three functions (Cost, Revenue and Profit), we can answer many questions. For example, what does it take
for the business to break even? This means, when will the profit be zero? So to answer this question we need to solve the
equation \(P(x)=0\). In other words, we need to find the zeros of the Profit function. Using the formula we have found, we get
\begin{eqnarray*}
2x-100,\!000&=&0\\
2x&=&100,\!000\\
x&=&50,\!000
\end{eqnarray*}
So the company will break even when \(x=50,\!000\), that is, when it has sold 50,000 units. This value of \(x\) is also called
the \textit{break-even point}.
\end{example}
\subsection{Physics applications}
Many Physics laws can be efficiently formulated using functions. A typical example is the description
of physical quantities that are changing with time. In this case, we use the variable \(t\) (for time) to describe
the input variable that produces as output the physical quantity.
\begin{example}
Suppose a circular oil slick in the ocean is expanding, so that the radius is increasing with time. After
some measurements, we determine that the radius \(r\) is changing according to the function
\[r(t)=3.5\sqrt{t}\]
where \(t\) is measured in minutes and \(r(t)\) in feet.
We want to determine the area of the oil slick as a function of time, including the right units.
Recall that the area of a circle is \(A=\pi r^2\). Since \(r\) is measured in feet, the units for \(A\) will be \(\mbox{ft}^2\).
So we find
\[
\begin{array}{rcll}
A(t) &=& \pi (3.5 \sqrt{t})^2 &\mbox{Substitute the expression for \(r\)}\\[2ex]
A(t)&=& \pi (3.5)^2 (\sqrt{t})^2 & \\[2ex]
A(t)&=&12.25 \pi t \mbox{ ft}^2& \mbox{Simplify}
\end{array}
\]
So the function \(A(t) = 12.25 t \mbox{ ft}^2\) expresses the area of the oil slick as a function of time.
\end{example}
Problems
\problem
The width of a rectangle is \(W=8\) inches. Express the area of the rectangle as a
function of its perimeter.
\begin{sol}
The formulas for the area and perimeter of a rectangle are
\[
\begin{array}{cc}
A=LW, & P=2L+2W
\end{array}
\]
where \(L\) is the length and \(W\) is the width.
We want to eliminate \(L\) and \(W\) in the formula \(A=LW\) and have just \(P\)
and numbers replace them.
We are given that \(W=8\), so we substitute this value:
\begin{eqnarray*}
A&=&8L\\
P&=&2L+2(8)\\
P&=&2L+16
\end{eqnarray*}
Now solve the perimeter equation for \(L\):
\[
\begin{array}{rcll}
P & = &2L+16 & \mbox{The equation for the perimeter}\\[1ex]
P\color{red}{-16} & = & 2L+16 \color{red}{-16} & \mbox{Subtract 16 from both sides}\\[1ex]
P-16 & = & 2L & \mbox{Simplify}\\[1ex]
\displaystyle{\frac{P-16}{\color{red}{2}}} & = &
\displaystyle{\frac{2L}{\color{red}{2}}} & \mbox{Divide both sides by 2}\\[1ex]
L & = & \displaystyle{\frac{P-16}{2}} & \mbox{Simplify, and switch sides}
\end{array}
\]
Now we substitute what we found in the equation for the area:
\[
\begin{array}{rcll}
A & = &8L & \mbox{The equation for the area}\\[1ex]
A & = & \displaystyle{8\left(\frac{P-16}{2}\right)} & \mbox{Substitute the expression for \(L\)}\\[1ex]
A & = & 4(P-16) & \mbox{Simplify}\\[1ex]
A & = & 4P-64
\end{array}
\]
The expression
\[A=4P-64\]
gives the area as a function of the perimeter.
\end{sol} \mproblem
The length of a rectangle is \(L=5\) cm. Express the area of the rectangle as a function of its perimeter.
\problem
Express the area of a square as a function of its perimeter.
\begin{sol}
The perimeter of a square is
\[P=4x\]
where \(x\) is the length of the side of the square. Solving this for \(x\) we find
\[x=\frac{P}{4}\]
Now substitute this in the formula for the area of a square, \(A=x^2\),
because we want to eliminate \(x\) and rewrite \(A\) in terms of \(P\) and numbers:
\[
\begin{array}{rcll}
A & = & x^2 \\[1ex]
& = & \displaystyle{\left(\frac{P}{4}\right)^2}\\[1ex]
& = & \displaystyle{\frac{P^2}{16}}
\end{array}
\]
So the formula
\[A=\frac{P^2}{16}\]
expresses the area of a square as a function of its perimeter.
\end{sol}
\mproblem
Express the perimeter of a square as a function of its area.
\problem
A company incurs a fixed cost of \$75,000 to produce a certain product, and then it spends \$150
for each unit it manufactures. The selling price of one unit will be \$250.
\begin{enumerate}
\item Find the cost \(C(x)\) as a function of the number \(x\) of items produced.
\item Find the revenue \(R(x)\) as a function of \(x\).
\item Find the profit \(P(x)\) as a function of \(x\).
\item How many items of the product need to be produced and sold to get a profit of \$375,000?
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
The cost \(C(x)\) is given by the fixed cost, plus the the cost incurred in producing \(x\) items. So the cost function is
\[C(x)=150x + 75,\!000\].
\item
Since one unit will be sold for \$250, the revenue
from selling \(x\) units will be
\[R(x)=250x\].
\item
The profit is given by the revenue minus the cost. So the profit is
\[
\begin{array}{rcll}
P(x) & = & R(x)-C(x)\\[1ex]
& = & 250x-(150x + 75,\!000)\\[1ex]
& = & 250x-150x-75,\!000\\[1ex]
& = & 100x-75,\!000
\end{array}
\]
and so the formula
\[P(x)=100x-75,\!000\]
expresses the profit \(P(x)\) as a function of the number \(x\) of units produced and sold.
\item
We need to solve the equation \(P(x)=375,\!000\) for \(x\).
Using the formula for \(P(x)\) found in part c., we get:
\[
\begin{array}{rcll}
100x-75,\!000 & = & 375,\!000\\[1ex]
100x -75,\!000 \color{red}{+75,\!000} & = & 375,\!000 \color{red}{+75,\! 000} &
\mbox{Add 75,000 to both sides}\\[1ex]
100x & = & 450,\!000 & \mbox{Simplify}\\[1ex]
\displaystyle{\frac{100x}{\color{red}{100}}} & = & \displaystyle{\frac{450,\!000}{\color{red}{100}}} &
\mbox{Divide by 100}\\[1ex]
x & = & 4,\!500 & \mbox{Simplify, and find the answer}
\end{array}
\]
\end{enumerate}
\end{sol} \mproblem
The fixed cost for a shoe manufacturer is \$25,000, and the cost
for each pair of shoes produced and sold is \$35. A pair of shoes will be sold for
\$80.
\begin{enumerate}
\item
Find the cost \(C(x)\) as a function of the number \(x\) of pairs of shoes produced and sold.
\item
Find the revenue \(R(x)\) as a function of \(x\).
\item
Find the profit \(P(x)\) as a function of \(x\).
\item
How many pairs of shoes will need to be produced and sold to get a profit of \$65,000?
\end{enumerate}
\problem
A spherical balloon is inflating in such a way that the radius is changing with time according to the function
\[r(t)=1.75 t^2,\]
where \(t\) is in seconds and \(r\) in feet.
Find the volume of the balloon as a function of time,including the right units for the volume.
\begin{sol}
The volume of a sphere of radius \(r\) is
\[V=\frac{4}{3}\pi r^3.\] Since \(r\) is in feet, the units for the volume will be \(\mbox{ft}^3\).
So substituting the expression for \(r\) we find
\[
\begin{array}{rcll}
V(t)&=&\displaystyle \frac{4}{3} \pi (1.75 t^2)^3\\[2ex]
V(t)&=& 22.45 t^6 \mbox{ ft}^3. &
\end{array}
\]
\end{sol} \mproblem
A spherical balloon is being deflated, and the radius is changing according to the function
\[r(t) =\frac{2.8}{t+1},\]
where \(t\) is in seconds, and \(r\) is in \(\mbox{cm}\).
Find the volume of the balloon as a function of time, including the right units.