Exercise 1.2.20

Question

The monthly cost of driving a car depends on the number of miles driven. Lynn found that in May it cost her \$380 to drive 480 mi and in June it cost her \$460 to drive 800 mi. 
    \begin{enumerate}[label=(\alph*)]    
        \item Express the monthly cost $C$ as a function of the distance driven $d$, assuming that a linear relationship gives a suitable model.
        \item Use part (a) to predict the cost of driving 1500 miles per month.
        \item Draw the graph of the linear function. What does the slope represent?
        \item What does the $C$-intercept represent?
        \item Why does a linear function give a suitable model in this situation?        
    \end{enumerate}

Khagan Gulusoy · Accepted Answer

\begin{enumerate}[label=(\alph*)]

\item Recall that for any linear function $f:X 	o \mathbb{R}$ and for two points $(x_1,y_1), \ (x_2,y_2)$ belonging to its graph, this formula is true:
    \begin{equation}\label{lineq}
        \forall x\in X:\quad \frac{x-x_1}{x_2-x_1}=\frac{f(x)-y_1}{y_2-y_1},
    \end{equation}
    Since points $(480,380), \ (800,460)$ belong to the graph of $C,$ the statement ef{lineq} translates to:
    $$\frac{d-480}{800-480}=\frac{C(d)-380}{460-380}.$$
    $$\frac{d-480}{320}=\frac{C(d)-380}{80}.$$
    $$\frac{d}{4}-120=C(d)-380.$$
    $$C(d)=\frac{d}{4}+260.$$
    Thus, we define the function $m$ as follows:
    $$\forall d\in [0,+\infty):\quad C(d)=\frac{d}{4}+260.$$

\item For $d=1500\mathrm{~mi}$, we have:
    $$C(1500)=\frac{1500}{4}+260=635\$.$$

\item The slope of the graph of the function $C$ is equal to $\dfrac{1}{4}.$ This is the measure of how much the cost of driving a car changes as the distance traveled increases. For instance, if you drive an additional mile, the cost of your car trip will go up by a quarter of a dollar.
    %For example, if the distance traveled by car increases by $1$ mi, the cost will increase by $\dfrac{1}{4}\$.$
    \begin{figure}[H] \centering
    \begin{tikzpicture}
    \begin{axis}[axis lines=left, xmin=0, xmax=1600, ymax=700, ymin=0, xlabel={$d$}, ylabel={$C(d)$}, ytick={0,260,380,460}, xtick={0,480,800},x label style = {at={(ticklabel cs:0.9)},anchor=south,above=10mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=0:1600, samples=2, mark=none, color=violet, ultra thick] {x/4+260};
         \addplot[mark=*, mark options={scale=1, fill=violet}](0,260);
         \addplot[mark=*, mark options={scale=1, fill=violet}](480,380);
         \addplot[mark=*, mark options={scale=1, fill=violet}](800,460);
         \addplot +[dashed,green!60!black,mark=none] coordinates {(480,0) (480,380) (0,380)};
         \addplot +[dashed,blue,mark=none] coordinates {(800,0) (800,460) (0,460)};
    \end{axis}    
    \end{tikzpicture}
    \end{figure}

\item The $C$-intercept of $260$ indicates the standing costs of a car (e.g. expenses such as insurance, registration, and maintenance, even if the car is not being driven). In economics, this quantity is often described as \emph{fixed costs} of production.
    %The $C$-intercept $260$ represents the cost of driving a car when the car did not move.

\item In the given exercise, a linear function is an appropriate model as the usage of oil is directly proportional to the distance traveled by car and the cost of driving is directly proportional to the use of oil. Also, just having a car without driving is associated with costs which is why the nonzero intercept is plausible.
    %Because the usage of oil linearly depends on the distance traveled by car and the cost of driving linearly depends on the usage of oil, the linear function gives a suitable model in the situation described in the exercise.
    
\end{enumerate}

Exercise 1.2.20

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Exercise 1.2.20

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