Show that the surface area of a closed cuboid with square base and given volume is minimum, when it is a cube.

Let us assume that the square base has side x and height h.

Let us assume that given volume of the cuboid is V.

∴ V = x²h

⇒ h = V/(x²)

Now surface area of the cuboid is-

S = 2x² + 4xh

= 2x² + 4x[V/(x²)]

= 2x² + 4(V/x)

On differentiating S, we get-

To find critical points,

Taking second derivative of S, we get-

Hence, is a point of minimum.

Thus, the surface area and volume is minimum when h = x which implies that cuboid is a cube.