Answer to Question #271603 in Calculus for keke

Question

A box with a square base is taller than it is width. In order to send the box through the

PH mail, the height of the box and the perimeter of the base can sum to no more than

108 in. What is the maximum volume for such a box? 

Answer 1

We can write the perimeter as $4l+h$ and this is equal to 108 inches.

where l = length of base; and

h = height

So,

$\begin{gathered}4l+h = 108\\h=108-4l\end{gathered}$

The volume is $l^2(h)$

Plugging the expression for h, we have:

$\begin{gathered}V=l^2 h\\V=l^2(108-4l)\\V=108l^2-4l^3\end{gathered}$

The max volume can be found when this differentiated is equal to 0, or dV/dx=0

$\begin{gathered}V=108l^2-4l^3\\\frac{dV}{dl}=216l- 12l^2=0\\12l(18-l)=0\\ 12l = 0,\ 18-l = 0\\l=0,18\end{gathered}$

l can't be 0, so we take l = 18.

Thus, $h=108-4l = 108-4(18)=36.$

Max Volume = $l^2h=(18)^2 (36)=11664 \ in³$