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