Given the perimeter of the box and the total surface area what is the maximum volume of the box?

Answer 1

With a fixed surface area you can actually maximize the volume of the box without using multi-variable calculus. Here's how:

Suppose the dimensions are x, y, z with 2(xy + yz + zx) = C for some constant C. This implies xy + yz + zx = C1 for a different constant C1. We will use what mathematicians call the AM-GM inequality (Arithmetic mean - geometric mean inequality) which says that, for some positive numbers, the arithmetic mean is always greater than or equal to the geometric mean, with equality occurring iff all the numbers are equal.

The AM-GM inequality says that

(xy + yz + zx)/3 >= (xy*yz*zx)1/3 (couldn't do cube root)

((xy + yz + zx)/3)3/2 >= xyz

The left side is equal to a constant (since it has the expression for surface area), so the maximal value of xyz (the volume) is equal to that constant. This happens when x = y = z, i.e. the box is cube-shaped.