Question

A box with a square base is wider than it is tall. In order to send the box through the U.S. mail, the width of the box and the perimeter of one of the (nonsquare) sides of the box can sum to no more than 117 in. What is the maximum volume for such a box? Maximum Volume =

Answer 1

Answer:

Maximum Volume = 29,659.5 cubic inches

Step-by-step explanation:

We can write the perimeter as 2*(x+h) and this is equal to 117

So

$2(x+h) = 117\\x+h=58.5\\h=58.5-x$

The volume is  $x^2(h)$

Plugging the expression for h, we have:

$V=x^2 h\\V=x^2(58.5-x)\\V=58.5x^2-x^3$

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

$V=58.5x^2-x^3\\\frac{dV}{dx}=117x-3x^2=0\\3x(39-x)=0\\x=0,39$

x can't be 0, so we take x = 39

and thus h=58.5-x = 58.5-39=19.5

Max Volume = $x^2h=(39)^2 (19.5)=29,659.5$