Calculus
posted by kieran .
A closed box is to be rectangular solid with a square base. If the volume is 32in^3, determine the dimensions for which the surface area is minimum.

make a sketch
let the base be x by x, and the height be y
so x^2 y = 32
y = 32/x^2
SA = 2x^2 + 4xy
= 2x^2 + 4x(32/x^2)
= 2x^2 + 128/x
d(SA)/dx = 4x  128/x^2 = 0 for a min of SA
4x = 128/x^2
4x^3 = 128
x^3 = 32
x = 32^(1/3) or appr 3.175
y = 32/3.175^2
Well , what do you know, it happens to be a perfect cube. 
60.48