The Volume of a box with a square base ##x## by ##x## cm and height ##h## cm is ##V=x^2h##
The amount of material used is directly proportional to the surface area, so we will minimize the amount of material by minimizing the surface area.
The surface area of the box described is ##A=x^2 +4xh##
We need ##A## as a function of ##x## alone, so we’ll use the fact that ##V=x^2h = 32,000## cm^3
which gives us ##h = (32,000)/x^2##, so the area becomes:
##A=x^2 +4x((32,000)/x^2) = x^2 +(128,000)/x##
We want to minimize ##A##, so
##A’ = 2x-(128,000)/x^2 = 0## when ##(2x^3-128,000)/x^2 = 0##
Which occurs when ##x^3 – 64,000 = 0## or ##x=40##
The only critical number is ##x=40## cm.
The second derivative test verifies that ##A## has a minimum at this critical number: ##A” = 2+(256,000)/x^3## which is positive at ##x = 40##.
The box should have base 40 cm by 40 cm and height 20 cm.
(use ##h = (32,000)/x^2## and ##x=40##)
