(i) Using similar triangles, \(\frac{6}{12} = \frac{r}{12-h}\). Solving for \(h\), we get \(h = 12 - 2r\).

The volume of the cylinder is \(V = \pi r^2 h\). Substituting \(h = 12 - 2r\), we have:

\(V = \pi r^2 (12 - 2r) = 12\pi r^2 - 2\pi r^3\).

(ii) Differentiate \(V\) with respect to \(r\):

\(\frac{dV}{dr} = 24\pi r - 6\pi r^2\).

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

\(24\pi r - 6\pi r^2 = 0\).

Factor out \(6\pi r\):

\(6\pi r (4 - r) = 0\).

Thus, \(r = 0\) or \(r = 4\). Since \(r = 0\) is not feasible, \(r = 4\).

Substitute \(r = 4\) into the volume formula:

\(V = 12\pi (4)^2 - 2\pi (4)^3 = 64\pi\).