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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 minus 2x)2 = a2 minus 4ax + 4x2

therefore Volume of box, V = a2x minus 4ax2 + 4x2

dV/dx = a2 minus 8ax + 12x2

At turning point, dV/dx = 0

therefore a2 minus 8ax + 12x2 = 0

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

therefore 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 = minus8a + 24x.

When x = a/6, d2V/dx2 = minus4a < 0 implies 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 minus 2x)

Solving Abase = Asides, (a minus 2x)2 = 4x(a minus 2x)

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

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

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

therefore 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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