(i) The surface area \(A\) of a closed cylinder is given by \(A = 2\pi r^2 + 2\pi rh\).

The volume \(V\) of the cylinder is \(\pi r^2 h = 1000\), so \(h = \frac{1000}{\pi r^2}\).

Substitute \(h\) into the surface area formula:

\(A = 2\pi r^2 + 2\pi r \left( \frac{1000}{\pi r^2} \right) = 2\pi r^2 + \frac{2000}{r}\).

(ii) To find the stationary value, differentiate \(A\) with respect to \(r\):

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

Set \(\frac{dA}{dr} = 0\) to find the stationary point:

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

Solving for \(r\), we get \(r^3 = \frac{2000}{4\pi}\).

\(r = \sqrt[3]{\frac{2000}{4\pi}} \approx 5.4\) cm.

To verify it's a minimum, check the second derivative:

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

Since \(\frac{d^2A}{dr^2} > 0\), \(A\) is minimized, confirming the flask is most efficient at \(r = 5.4\) cm.