Cylinder inscribed in a cone

The second standard problem: to find, among all cylinders inscribed in a cone, the one of maximum volume. The result ($4/9$ of the volume of the cone) is surprisingly small.

Example — Volume of the cylinder inscribed in a cone

A cone has base radius $R$ and height $H$. Find the inscribed cylinder (axis coinciding with that of the cone) of maximum volume.

Axial section: the cylinder $(r,h)$ is inscribed in the triangle representing the cone $(R,H)$.

Solution

By similarity, if $r$ is the radius of the cylinder and $h$ its height: $\frac{r}{R} = \frac{H-h}{H} \iff h = H\bigl(1-r/R\bigr) = H-\frac{H}{R}r.$ Volume: $V(r)=\pi r^2\,h = \pi r^2 H\bigl(1-r/R\bigr) = \pi H\bigl(r^2 - r^3/R\bigr)$, with $0<r<R$. $V'(r) = \pi H\bigl(2r - 3r^2/R\bigr) = \pi H\,r\bigl(2 - 3r/R\bigr) = 0 \iff r=\frac{2R}{3}.$ Then $h=H(1-2/3)=H/3$. The optimal cylinder takes up $2R/3$ in radius and $H/3$ in height. $V_{\max} = \pi (2R/3)^2 (H/3) = \frac{4\pi R^2 H}{27} = \frac{4}{9}\cdot\frac{\pi R^2 H}{3} = \boxed{\frac{4}{9}\,V_{\text{cono}}}.$ The maximum cylinder occupies $4/9$ of the volume of the cone that contains it: less than half, a counterintuitive result.

Links

Topics: Derivatives
Concepts: Objective function · Maximum and minimum
Methods: Maximum minimum via derivative
Skills: Synthetic geometry · Modelling