4. In a given circular cone, with height $H$ and base radius $R$, a cylinder with maximum volume is inscribed. In this cylinder, a cone is inscribed, and in that cone, another cylinder with maximum volume is inscribed, etc. (the bases of the cylinders and cones are in the same plane). Find the sum of the volumes of the inscribed cylinders.
Solution. Let's denote by $x$ the radius of the base of the inscribed cylinder, and by $y$ its height. It is clear that $\triangle A B C \sim \triangle A^{\prime} B^{\prime} C$ (see the diagram; both triangles have one common acute angle and both are right triangles). From their similarity, we obtain

$$
x=\frac{R}{H}(H-y)
$$

We are looking for the cylinder that has the maximum volume $V=\pi x^{2} y$, i.e., we will find for which $y$ the function

$$
\Phi(y)=\frac{R^{2}}{H^{2}}(H-y)^{2} y
$$

has the maximum value. From

$$
\Phi^{\prime}(y)=\pi \frac{R^{2}}{H^{2}}\left[2(H-y)(-1) y+(H-y)^{2}\right]=0
$$

we obtain the equation

$$
3 y^{2}-4 H y+H^{2}=0
$$

whose solutions are $y_{1}=H$ and $y_{2}=\frac{H}{3}$. Due to

$$
\Phi^{\prime \prime}(y)=\pi \frac{R^{2}}{H^{2}}\left[2(-1)(-1) y+2(H-y)(-1)+2(H-y)(-1)=\pi \frac{R^{2}}{H^{2}}(-4 H+6 y)\right.
$$

we have

$$
\Phi^{\prime \prime}\left(y_{1}\right)=\pi \frac{R^{2}}{H^{2}} 2 H>0
$$

which means that $\Phi\left(y_{1}\right)$ is a minimum of the function $\Phi(y)$ (the cylinder degenerates to the height of the cone), and $\Phi^{\prime \prime}\left(y_{2}\right)=\pi \frac{R^{2}}{H^{2}}(-2 H)<0$, which means that for $y=\frac{H}{3}$, the maximum of $\Phi(y)$ is obtained. For this value of $y$, from equation (1) we get $x=\frac{2}{3} R$ and the maximum volume will be

$$
\Phi\left(y_{1}\right)=4 \pi \frac{R^{2} H}{27}
$$

If we now inscribe a cone in the obtained cylinder (the height of this cone is the height of the cylinder, $\frac{H}{3}$, and the radius of the base of the cone is the radius of the base of the cylinder, $\frac{2}{3} R$) and inscribe a cylinder with maximum volume in this cone, then the height of the newly inscribed cylinder will be $\frac{1}{3}\left(\frac{H}{3}\right)=\frac{1}{3^{2}} H$, and the radius of its base will be

$$
\frac{2}{3} \cdot \frac{2}{3} R=\left(\frac{2}{3}\right)^{2} R
$$

while its volume is $\pi R^{2} H\left(\frac{2}{3}\right)^{4}\left(\frac{1}{3}\right)^{2}$; the third inscribed cylinder will have a height of $\left(\frac{1}{3}\right)^{3} H$, a base radius of $\left(\frac{2}{3}\right)^{3} R$, and a volume of $\pi R^{2} H\left(\frac{2}{3}\right)^{6}\left(\frac{1}{3}\right)^{3}$, and similarly for each subsequent inscribed cylinder. In this way, we obtain a sequence of numbers (volumes) that represents a geometric series with common ratio $q=\frac{4}{27}<1$, and therefore the sum of the infinite geometric series will be

$$
\frac{4 \pi R^{2} H \frac{1}{27}}{1-\frac{4}{27}}=\frac{4}{23} \pi R^{2} H
$$