Let's inscribe a right circular cylinder of the largest possible volume in a given right circular cone. What percentage of the cone's volume is the volume of the cylinder? (The base of the inscribed cylinder lies in the base of the cone, and the other circle of the cylinder is a plane section of the cone.)
The figure shows a cross-section of a cone and a cylinder through their common axis.

![](https://cdn.mathpix.com/cropped/2024_05_02_682052b72d2e0233547eg-1.jpg?height=493&width=504&top_left_y=239&top_left_x=794)

If the radius of the cylinder is denoted by $r$ and its height by $m$, the volume is

$$
V=r^{2} \pi m
$$

From the similar triangles in the figure:

$$
\frac{m}{M}=\frac{R-r}{R}
$$

Solving for $m$ and substituting it into the volume formula:

$$
V=\frac{M \pi}{R} r^{2}(R-r)=\frac{4 M \pi}{R} \cdot \frac{r}{2} \cdot \frac{r}{2}(R-r)
$$

In the product, the sum of the factors depending on $r$ is $\frac{r}{2}+\frac{r}{2}+R-r=R$, which is constant. Therefore, the volume will be maximized when the factors are equal:

$$
\frac{r}{2}=R-r
$$

which gives

$$
r=\frac{2}{3} R
$$

The maximum volume is

$$
V_{\max }=\frac{4}{27} R^{2} \pi M
$$

This is $\frac{4}{9}$ of the volume of the cone, which is $\frac{R^{2} \pi M}{3}$, or $100 \cdot \frac{4}{9}=44 \frac{4}{9} \%$.

With this, we have solved the problem.

(Bartha László, Balassagyarmat, Balassa g. III. o. t.)