Question

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

Answer 1

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

h = V/x² ----- (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

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:

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

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:

Now let us find the value of

so the function S is minimum at x = \(\sqrt[3]{V}\)

Substituting x in equation (1)

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