Answer: \(r=\sqrt[3]{\dfrac{660}{4\pi}}\approx3.74\text{ cm}\), and this gives a minimum.

(a) The total surface area of a closed cylinder is

\(A=2\pi r^2+2\pi rh,\)

where \(2\pi r^2\) is the area of the two circular ends and \(2\pi rh\) is the curved surface area.

The volume is

\(\pi r^2h=330.\)

So

\(h=\dfrac{330}{\pi r^2}.\)

Substitute this into the expression for \(A\):

\(A=2\pi r^2+2\pi r\left(\dfrac{330}{\pi r^2}\right).\)

Simplifying,

\(A=2\pi r^2+\dfrac{660}{r},\)

as required.

(b) Differentiate with respect to \(r\):

\(\dfrac{\mathrm dA}{\mathrm dr}=4\pi r-\dfrac{660}{r^2}.\)

At a stationary value,

\(\dfrac{\mathrm dA}{\mathrm dr}=0.\)

So

\(4\pi r-\dfrac{660}{r^2}=0.\)

Multiplying by \(r^2\),

\(4\pi r^3=660.\)

Thus

\(r^3=\dfrac{660}{4\pi},\)

and

\(r=\sqrt[3]{\dfrac{660}{4\pi}}=3.744\ldots\text{ cm}.\)

So

\(r\approx3.74\text{ cm}.\)

To show that this stationary value is a minimum, differentiate again:

\(\dfrac{\mathrm d^2A}{\mathrm dr^2}=4\pi+\dfrac{1320}{r^3}.\)

For \(r\gt0\), both terms are positive, so

\(\dfrac{\mathrm d^2A}{\mathrm dr^2}\gt0.\)

Therefore the stationary value is a minimum.