Find the height and the radius of the base of the cylinder of the largest volume inscribed in a sphere of radius $R$.

#
Let $h$ be the height of the cylinder, $r$ be the radius of its base (Fig.1). Consider the section of the sphere by a plane passing through its center $O$ (Fig.2). In the section, we get a circle of radius $R$, in which a rectangle $A B C D$ is inscribed with sides $A D=B C=2 r, A B=C D=h$ and diagonal $A C=2 R$, and the center $O$ of the circle coincides with the center of the rectangle $A B C D$. From the right triangle $A C D$, we find that

$$
4 r 2=A D 2=A C 2-C D 2=4 R 2-h 2
$$

Let $V(h)$ be the volume of the cylinder. Then

$$
V(h)=\pi r 2 h=\frac{\mathbf{1}}{\mathbf{4}} \pi h(4 R 2-h 2)
$$

Thus, the problem reduces to finding the maximum value of the function $V(h)=\frac{1}{4} \pi h(4 R 2-h 2)$ on the interval $(0 ; 2 R)$. Solving the equation $V^{\prime}(h)=0$, we will find the critical points of the function $V(h)$. We will consider only those of them that belong to the interval $(0 ; 2 R)$.

$$
V^{\prime}(h)=\left(\frac{1}{4} \pi h(4 R 2-h 2)\right)^{\prime}=\frac{1}{4} \pi(4 R 2 h-h 3)^{\prime}=\pi\left(R 2-\frac{3}{4} h 2\right) .
$$

The interval $(0 ; 2 R)$ contains the only root of this equation $h=\frac{2 \cdot R}{\sqrt{3}}$. When passing through the point $h=\frac{2 R}{\sqrt{3}}$, the derivative changes sign from plus to minus. This means that on the interval $(0 ; \sqrt{3})$ the function $V(h)$ increases, and on the interval $\left(\frac{2 R}{\sqrt{3}} ; 2 R\right)$ it decreases. Therefore, when $h=\frac{2 R}{\sqrt{3}}$ the volume of the cylinder is the largest. In this case,

$$
r=\frac{1}{2} \sqrt{4 R^{2}-h^{2}}=\frac{1}{2} \sqrt{4 R^{2}-\frac{4}{3} R^{2}}=R \sqrt{\frac{2}{3}}
$$

## Answer

$\frac{2 R}{\sqrt{3}} ; R \sqrt{\frac{2}{3}}$