Solution
= Solution
The curved <surface area> excludes the two end disks, so $A=2\pi rx=2\pi\tan\alpha\,x(h-x)$. Completing the square gives
$$
A=2\pi\tan\alpha\left[\frac{h^2}{4}-\left(x-\frac h2\right)^2\right].
$$
Thus
$$
\boxed{x=\frac h2,\qquad A_{\max}=\frac{\pi h^2\tan\alpha}{2}.}
$$
It differs from the volume optimum because the objective contains one power of the radius rather than two.