Given,

• Closed cuboid has square base.

• The volume of the cuboid is given.

• Surface area is minimum.

Let us consider,

• The side of the square base be ‘x’.

• The height of the cuboid be ‘h’.

• The given volume, V = x²h

----- (1)

Consider the surface area of the cuboid,

Surface Area =

2(Area of the square base) + 4(areas of the rectangular sides)

S = 2x² + 4xh

Now substitute (1) in the Surface Area formula

----- (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function S(x) has a maximum/minimum at a point c then S’(c) = 0.

Differentiating the equation (2) with respect to x:

[Since and ]

------- (3)

To find the critical point, we need to equate equation (3) to zero.

x³ = V

Now to check if this critical point will determine the minimum surface area, we need to check with second differential which needs to be positive.

Consider differentiating the equation (3) with x:

----- (4)

[Since and ]

Now let us find the value of

As , so the function S is minimum at

Substituting x in equation (1)

h = x

As S is minimum and h = x, this means that the cuboid is a cube.