Answer: \(r=\sqrt[3]{\frac{500}{\pi}}\), approximately \(5.42\text{ cm}\).

We start with the main method. Model the quantity to be optimised as a function of one variable, then differentiate to locate the stationary value.

(a) The volume of the cylinder is

\(\pi r^2h=1000.\)

So

\(h=\frac{1000}{\pi r^2}.\)

The total surface area of a closed cylinder is

\(S=2\pi r^2+2\pi rh.\)

Substitute \(h=\frac{1000}{\pi r^2}\):

\(S=2\pi r^2+2\pi r\left(\frac{1000}{\pi r^2}\right).\)

Therefore

\(S=2\pi r^2+\frac{2000}{r}.\)

(b) Differentiate:

\(\frac{\mathrm dS}{\mathrm dr}=4\pi r-\frac{2000}{r^2}.\)

At a stationary value,

\(4\pi r-\frac{2000}{r^2}=0.\)

Thus

\(4\pi r^3=2000,\)

so

\(r^3=\frac{500}{\pi}.\)

Hence

\(r=\sqrt[3]{\frac{500}{\pi}}.\)

To check that this gives a minimum,

\(\frac{\mathrm d^2S}{\mathrm dr^2}=4\pi+\frac{4000}{r^3}.\)

For \(r\gt0\), this is positive, so the stationary value is a minimum.

Therefore

\(r=\sqrt[3]{\frac{500}{\pi}}\approx5.42\text{ cm}.\)

This completes the solution and gives the required result.