A squarebottomed box with no top has a fixed volume V cubic

Solution

Since it is a square-bottomed box, let length and breadth be x.

Let height be y.

Volume = V = x²y. y = V/x²

Surface Area = S = x²+4xy = x²+4V/x = f(x)

We have to minimize surface area so,

f\'(x) = 2x - 4V/x² = 0.    x³ = 2V x = ³2V

y = ³V/4

Therefore for minimum surface area, length and breadth = ³2V and height = ³V/4.