8,9

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

#
Let $P$ be the vertex of the cone, $h$ its height, and $r$ the radius of the base (Fig.1). Consider the section of the sphere by a plane passing through its center $O$. In the section, a circle of radius $R$ (Fig.2) is obtained, in which an isosceles triangle $A B P$ is inscribed with vertex $P$, base $A B=2 r$, and height $P M=h$. Extend the height $P M$ until it intersects the circle at point $K$. Then $P B K$ is a right triangle, and $B M$ is the height drawn from the right angle vertex. Therefore,

$$
\text { BM2 }=\text { PM } \cdot \text { KM, or } r 2=h(2 R-h)
$$

$$
V(h)=\frac{1}{3} \pi r 2 h=\frac{1}{3} \pi h 2(2 R-h)
$$

Thus, the problem reduces to finding the maximum value of the function $V(h)=\frac{1}{3} \pi h 2(2 R-h)$ on the interval $(0 ; 2 R)$.

## First method.

By solving the equation $V^{\prime}(h)=0$, we will find the critical points of the function $V(h)$. Consider only those that belong to the interval $(0 ; 2 R)$

$$
V^{\prime}(h)=\left(\frac{1}{3} \pi(2 R h 2-h 3)\right)^{\prime}=\frac{1}{3} \pi(4 R h-3 h 2)=\frac{1}{3} \pi h(4 R-3 h)=0
$$

The interval $(0 ; 2 R)$ contains only one root of this equation $h=\frac{4}{3} R$. As we pass through the point $h=\frac{4}{3} R$, the derivative changes sign from positive to negative. Therefore, on the interval $\left(0 ; \frac{4}{3} R\right)$ the function $V(h)$ is increasing, and on the interval $\left(\frac{4}{3} R ; 2 R\right)$ it is decreasing. Thus, at $h=\frac{4}{3} R$ the volume of the cone is the largest. In this case,

$$
r=\sqrt{h(2 R-h)}=\sqrt{\frac{4}{3} R\left(2 R-\frac{4}{3} R\right)}=\frac{2}{3} R \sqrt{2}
$$

## Second method.

Applying the Cauchy inequality to three numbers, we get that

$$
\begin{aligned}
V(h) & =\frac{1}{3} \pi h 2(2 R-h)=\frac{1}{3} \pi \cdot 4 \cdot \frac{1}{2} h \cdot \frac{1}{2} h \cdot(2 R-h) \leqslant \\
& \leqslant \frac{4}{3} \pi\left(\frac{\frac{1}{2} h+\frac{1}{2} h+(2 R-h)}{3}\right) 3=\frac{4}{3} \pi \cdot \frac{8 R}{27}=\frac{32 \pi}{81},
\end{aligned}
$$

with equality achieved if $\frac{1}{2} h=2 R-h$, i.e., at $h=\frac{4}{3} R$. Therefore, the maximum value of the cone's volume is achieved at $h=\frac{4}{3} R$. In this case,

$$
r=\sqrt{h(2 R-h)}=\sqrt{\frac{4}{3} R\left(2 R-\frac{4}{3} R\right)}=\frac{2}{3} R \sqrt{2}
$$

## Answer

$\frac{4}{3} R ; \frac{2}{3} R \sqrt{2}$.