## 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.

A_{base} = (`a` 2`x`)^{2} = `a`^{2} 4`ax` + 4`x`^{2}

Volume of box, V = `a`^{2}`x` 4`ax`^{2} + 4`x`^{2}

^{dV}/_{dx} = `a`^{2} 8`ax` + 12`x`^{2}

At turning point, ^{dV}/_{dx} = 0

`a`^{2} 8`ax` + 12`x`^{2} = 0

(`a` 2`x`)(`a` 6`x`) = 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.

^{d2V}/

_{dx2}= 8

`a`+ 24

`x`.

When `x` = `a`/6, ^{d2V}/_{dx2} = 4`a` < 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} = 4`x`(`a` 2`x`)

Solving A_{base} = A_{sides}, (`a` 2`x`)^{2} = 4`x`(`a` 2`x`)

(`a` 2`x`)^{2} 4`x`(`a` 2`x`) = 0

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

(`a` 2`x`)(`a` 6`x`) = 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.