4.4. Find the radius of the base and the height of a right circular cone of the smallest volume described around a sphere of radius $R$.


4.4. Let $S$ be the vertex of the cone. The intersection of the cone and one of the planes that contain $S$ and are perpendicular to the base of the cone is a triangle $SAB$. Let $O$ be the center of the inscribed circle in triangle $SAB$ and $D$ the point of tangency of the inscribed circle on $SA$. The line $SO$ contains the height $ST$ of triangle $SAB$ (since $SA=SB$). $AT=r, ST=H$,

$OD=OT=R$, where $r$ and $H$ denote the radius of the base and the height of the cone. Triangle $STA$ is similar to triangle $SOD$. Therefore, we have:

$$
\begin{gathered}
\frac{r}{R}=\frac{H}{\sqrt{H(H-2 R)}} \text {, hence } r^{2}=\frac{R^{2} H}{H-2 R}, \text { so } \\
V=\frac{\pi}{3} r^{2} H=\frac{\pi}{3} R^{2} \frac{H^{2}}{H-2 R} \text { and } V^{\prime}(H)=\frac{\pi}{3} R^{2} \frac{H(H-4 R)}{(H-2 R)^{2}}
\end{gathered}
$$

Thus, for $H<4R$ we have $V^{\prime}(H)<0$ and for $H>4R$ we have $V^{\prime}(H)>0$. This means the function $V(H)$ reaches a minimum at $H=4R$. Then $r=R \sqrt{2}$.