College Algebra, Chapter 4, Chapter Review, Section Review, Problem 22

Suppose that the box is to be constructed of thin plastic natural. It will have square ends and a rectangular top and back, with an open bottom and front. The total area of the four plastic sides is to be $1200$ in$^2$.

a.) Express the volume $V$ of the shelter as a function of the depth $x$.

b.) Draw the graph of $V$.

c.) What dimensions will maximize the volume of the shelter?



a.) If the total area of the four sides is $1200$ in$^2$, then


$
\begin{equation}
\begin{aligned}

1200 =& x^2 + x^2 + xy + xy
\\
\\
1200 =& 2x^2 + 2xy

\end{aligned}
\end{equation}
$


So $\displaystyle y = \frac{1200 - 2x^2}{2x}$

Then, recall that


$
\begin{equation}
\begin{aligned}

V =& x^2 y
\\
\\
V =& x^2 \left( \frac{1200 - 2x^2}{2x} \right)
\\
\\
V =& \frac{x}{2} (1200 - 2x^2)
\\
\\
V =& 600 x - x^3
\\
\\
V(x) =& 600 x - x^3

\end{aligned}
\end{equation}
$


b.)








Based from the graph, the volume is maximum when $x \approx 14$ in.. So, if $x \approx 14 $ in, then

$\displaystyle y \approx \frac{1200 - 2 (14)^2}{2(14)} \approx 28.86 in$

Therefore, the dimensions that will maximize the volume is

$x \approx 14$ in and $y \approx 28.86$ in