 ## Maximised Box

#### 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. Abase = (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 Abase = 0, so we need only consider the solution x = a/6.

d2V/dx2 = 8a + 24x.

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.

Asides = 4x(a 2x)

Solving Abase = Asides, (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 Abase = 0.

Hence Abase = Asides 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.

Note: Despite its apparent similarities with the last problem this is very difficult to prove.

Problem ID: 121 (May 2003)     Difficulty: 4 Star

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