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A square-based cuboid is to be formed from pieces of steel tubing and then wrapped in fabric. The total length of steel tubing used is $64 m$ .

An engineer draws a diagram to model the situation.

Cuboid diagram

The fabric costs $£ 7.50 m^{2}$ . What is the maximum cost of covering the cuboid?
Prove that $A$ achieves its maximum value when the cuboid is a cube.

Give your answer to (a) in £ below.

Hint

The total amount of tubing used in the engineer's diagram is

8 x + 4 y = 64

Hint

The total surface area of the cuboid is

A = 2 x^{2} + 4 x y

Solution

The total amount of tubing used in the engineer's diagram is

\begin{aligned} 8 x + 4 y & = 64 \\ 2 x + y & = 16 \\ y & = 16 - 2 x \end{aligned}

The total surface area of the cuboid is

\begin{aligned} A & = 2 x^{2} + 4 x y \\ = 2 x^{2} + 4 x (16 - 2 x) \\ = 2 x^{2} + 64 x - 8 x^{2} \\ = 64 x - 6 x^{2} \\ = - 6 [x^{2} - \frac{32}{3} x] \\ = - 6 [{(x - \frac{16}{3})}^{2} - \frac{256}{9}] \\ = \frac{512}{3} - 6 {(x - \frac{16}{3})}^{2} \end{aligned}

The maximum area is $\frac{512}{3}$ and so the maximum cost is

\frac{512}{3} \times £ 7.50 = £ 1280

This maximum occurs when $x = \frac{16}{3}$ , in which case

\begin{aligned} y & = 16 - 2 x \\ = 16 - \frac{32}{3} \\ = \frac{16}{3} \end{aligned}

which is to say, the cuboid is a cube.