mathschallenge.net

Problem

Four corners measuring x by x are removed from a sheet of material that measures a by a to make a square based open-top box.

Prove that the volume of the box is maximised iff the area of the base is equal to the area of the four sides.

Solution

Consider the diagram.

A_base = (a 2x)² = a² 4ax + 4x²

Volume of box, V = a²x 4ax² + 4x²

^dV/_dx = a² 8ax + 12x²

At turning point, ^dV/_dx = 0

a² 8ax + 12x² = 0

(a 2x)(a 6x) = 0

x = a/2, a/6.

Clearly x = a/2 is a trivial solution, as A_base = 0, so we need only consider the solution x = a/6.

When x = a/6, ^d2V/_dx² = 4a < 0 V is at a maximum value.

Note that it is not sufficient to show that the area of the base equals the area of the sides when x = a/6, as we are attempting to prove that the volume is maximised if and only if this condition is true.

A_sides = 4x(a 2x)

Solving A_base = A_sides, (a 2x)² = 4x(a 2x)

(a 2x)² 4x(a 2x) = 0

(a 2x)((a 2x) 4x) = 0

(a 2x)(a 6x) = 0

x = a/2, a/6.

We reject x = a/2, as A_base = 0.

Hence A_base = A_sides has a unique non-trivial solution, x = a/6, which is when the volume of the box is maximised.

Prove the same holds for an open-top box with a rectangular base.