A box has a bottom with one edge 2 times as long as the other. If the box has no top and the volume is fixed at , what dimensions minimize the surface area?

Answer 1

width of base --- x
length of base -- 2x
height --- k, where k is a constant

Volume = c = 2kx^2 , where c is a constant
k = c/(2x^2)

surface area = 2x^2 + 2kx + 4kx
= 2x^2 +4kx
= 2x^2 + 4x(c/(2x^2)
= 2x^2 + 2c/x

d(surface area)/dx = 4x -2c/x^2 = 0 for a min of SA
4x = 2c/x^2
2x^3 = c
x^3 = c/2
x = (c/2)^(1/3)
2x = 2(c/2)^(1/3)
k = c/(2(c/2)^(2/3)
= (1/2)c^(1/3)