= Solution
At fixed $h,\alpha$, the <derivative> is $V'=\pi\tan^2\alpha\,(h-x)(h-3x)$. It is positive for $0<x<h/3$ and negative for $h/3<x<h$. The endpoint volumes are zero. Thus the unique positive-volume global <maximum> is at
$$
\boxed{x=\frac h3.}
$$
Its value is $4\pi h^3\tan^2\alpha/27$, whereas the cone volume is $\pi h^3\tan^2\alpha/3$. Hence
$$
\boxed{\frac{V_{\rm cylinder}}{V_{\rm cone}}=\frac49.}
$$
This is the volume case of the <optimal cylinder inscribed in a cone>.
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