Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 11 / c

An axial section gives similar right triangles. At height $x$ above the base, the cone's remaining height is $h-x$, so the cylinder's radius is $r=(h-x)\tan \alpha$. Therefore its volume isAssume $h>0$ and $0<\alpha <\pi /2$ for a nondegenerate cone. The geometry and the dimensionless optimization curves are illustrated below.

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.

The curved surface area excludes the two end disks, so $A=2\pi rx=2\pi \tan \alpha x(h-x)$. Completing the square givesThusIt differs from the volume optimum because the objective contains one power of the radius rather than two.