Answer: \(S_{\min}\approx1340\text{ cm}^2\), minimum.

The volume is

\(\pi r^2h=1200\pi.\)

So

\(h=\frac{1200}{r^2}.\)

The total surface area of a closed cylinder is

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

Substitute \(h=\frac{1200}{r^2}\):

\(S=2\pi r^2+2\pi r\left(\frac{1200}{r^2}\right).\)

Hence

\(S=2\pi r^2+\frac{2400\pi}{r}.\)

Differentiate:

\(\frac{dS}{dr}=4\pi r-\frac{2400\pi}{r^2}.\)

For a stationary value, set \(\frac{dS}{dr}=0\):

\(4\pi r-\frac{2400\pi}{r^2}=0.\)

So

\(4r^3=2400,\)

and therefore

\(r=\sqrt[3]{600}\approx8.43.\)

Substituting this value into \(S\) gives

\(S\approx1340.\)

To determine the nature,

\(\frac{d^2S}{dr^2}=4\pi+\frac{4800\pi}{r^3}.\)

This is positive for \(r\gt 0\), so the stationary value is a minimum.

Thus

\(\boxed{S_{\min}\approx1340\text{ cm}^2}.\)