$\left[\begin{array}{l}[\text { Areas and volumes (extremum problems) }] \\ {[\underline{\text { Cone }}}\end{array}\right]$

A cone is described around a cube in such a way that four vertices of the cube lie in the plane of the base of the cone, and the other four vertices lie on its lateral surface. What is the smallest volume that such a cone can have if the edge of the cube is $a$?
Let the vertices $A, B, C, D$ of the cube $A B C D A 1 B 1 C 1 D 1$ (Fig.1) lie on the lateral surface of a cone with vertex $P$, and the vertices $A 1, B 1, C 1, D 1$ lie in the plane of the base ( $A A 1\|B B 1\| C C 1 \| D D 1$ ). We draw an axial section of the cone passing through the vertices $A$ and $C$ of the cube. We obtain an isosceles triangle $P M N$ (Fig.2), in which the rectangle $A C C 1 A 1$ is inscribed, with vertices $A$ and $C$ lying on the sides $P M$ and $P N$ respectively, and vertices $A 1$ and $C 1$ on the base $M N$, such that $C C 1=A A 1=a, A 1 C 1=A C=a \sqrt{2}$. Let $P O=h$ be the height of the cone, and $O M=O N=r$ be the radius of its base. From the similarity of triangles $A P C$ and $M P N$, it follows that

$$
\frac{P O}{M N}=\frac{P O-A A_{1}}{A C}, \text { or } \frac{h}{2 \varphi}=\frac{h-a}{\Delta \sqrt{2}}
$$

from which we find that $r=\frac{ah \sqrt{2}}{2(h-a)}$. Let $V(h)$ be the volume of the cone. Then

$$
V(h)=\frac{1}{3} \pi r 2 h=\frac{1}{6} \pi a 2 \cdot \frac{h^{3}}{(h-a)^{2}} .
$$

Thus, the problem reduces to finding the maximum value of the function $V(h)=\frac{1}{6} \pi a 2 \cdot \frac{h^{3}}{(h-a)^{2}}$ on the ray ( $a$; + $(\infty)$ ). Solving the equation $V^{\prime}(h)=0$, we find the critical points of the function $V(h)$. We consider only those that belong to the interval ( $a ;+\infty$ ).

$$
\begin{aligned}
& V^{\prime}(h)=\frac{1}{6} \pi a 22^{\left.\frac{h^{3}}{(h-\alpha)^{2}}\right)^{\prime}}=\frac{1}{6} \pi a 2^{\frac{3 h^{2}(h-\alpha)^{2}-2 h^{3}(h-\alpha)}{(h-\alpha)^{1}}}= \\
& =\frac{1}{6} \pi a 2^{\frac{h^{2}(h-\alpha)(3(h-\alpha)-2 h)}{(h-\alpha)^{1}}}=\frac{1}{6} \pi a 2^{\frac{h^{2}(h-3 a)}{(h-\Delta)^{3}}}=0 .
\end{aligned}
$$

The interval $(a ;+\infty)$ contains the only root of this equation $h=3 a$. As we pass through the point $h=$ $3 a$, the derivative changes sign from minus to plus. Therefore, on the interval $(0 ; 3 a)$ the function $V(h)$ decreases, and on the interval $(3 a ;+\infty)$ it increases. Thus, at $h=3 a$ the volume of the cone is the smallest. In this case,

$$
V(3 a)=\frac{1}{6} \pi a 2 \cdot \frac{(3 a)^{3}}{(3 a-a)^{2}}=\frac{9}{8} \pi a 3
$$

## Answer

$\frac{9}{8} \pi a 3$.