 Solution

At fixed $h,\alpha$, the derivative is $V^{\prime }=\pi {\tan }^2\alpha (h-x)(h-3x)$. 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 atIts value is $4\pi h^3{\tan }^2\alpha /27$, whereas the cone volume is $\pi h^3{\tan }^2\alpha /3$. HenceThis is the volume case of the optimal cylinder inscribed in a cone.