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.
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