2. A right cylinder with radius $r=\sqrt{39}$ and a right cone with radius $R=9$ have the same height $H=15$. At what non-zero height $x$, should the cylinder and the cone be intersected so that the newly formed cylinder with height $x$ and the intersected cone with height $x$ have equal volumes?
Solution. Let $y$ denote the radius of the smaller base of the truncated cone with height $x$. From the similarity of triangles $A O O^{\prime}$ and $A C C^{\prime}$, we get

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

For the volume of the truncated cone, we have

$$
\begin{aligned}
V_{k} & =\frac{R^{2} \pi H}{3}-\frac{y^{2} \pi(H-x)}{3} \\
& =\frac{\pi}{2}\left(R^{2} H-y^{2} H+y^{2} x\right)
\end{aligned}
$$

If we express $y$ from (1) and substitute it into (2)

![](https://cdn.mathpix.com/cropped/2024_06_05_d48bd9f430eb3870413eg-07.jpg?height=466&width=430&top_left_y=654&top_left_x=1035)
we get

$$
V_{k}=\frac{R^{2} \pi x}{3 H^{2}}\left(x^{2}-3 H x+3 H^{2}\right)
$$

By equating this volume to the volume of the cylinder with height $x$, we get the equation:

$$
r^{2} \pi x=\frac{R^{2} \pi x}{3 H^{2}}\left(x^{2}-3 H x+3 H^{2}\right)
$$

If we divide the equation by $x(x \neq 0)$ and substitute the given values for $r, H$, and $H$, we get

$$
x^{2}+45 x+350=0
$$

The solutions to this equation are 10 and 35, of which the latter is not considered because it exceeds the height of the given cylinder and cone.

Thus, the cylinder and cone should intersect at a height $x=10$.