A sphere with radius $r$ is inscribed in a cylinder, and the volume of the cylinder is $V_{1}$; the volume of a cone circumscribed around the sphere is $V_{2}$. What is the smallest value of the ratio $V_{2} / V_{1}$?
A sphere can be circumscribed by a cylinder in only one way, and the volume of this cylinder is constant, specifically expressed by the sphere's radius as $V_{1}=2 r^{3} \pi$.

The ratio $V_{2} / V_{1}$ will be the smallest when $V_{2}$ takes its smallest value.

Thus, among the cones that can be circumscribed around the sphere, the one with the smallest volume must be determined. Let the radius of the base of the cone be denoted by $R$ and its height by $m$; thus $V_{2}=\frac{R^{2} \pi m}{3}$ and

$$
\frac{V_{2}}{V_{1}}=\frac{\frac{R^{2} \pi m}{3}}{2 r^{3} \pi}=R^{2} m \cdot \frac{1}{6 r^{3}}
$$

Since $\frac{1}{6 r^{3}}$ is constant, we only need to examine when $R^{2} m$ is minimal.

Let's create a cross-section of the cone that passes through the center of the base and is perpendicular to the base plane.

![](https://cdn.mathpix.com/cropped/2024_05_02_6c3ba68e80639c8af263g-1.jpg?height=873&width=792&top_left_y=829&top_left_x=645)

Using the notation in the diagram, from the similar triangles $O E B$ and $B D C$,

$$
\frac{R}{m}=\frac{r}{\sqrt{(m-r)^{2}-r^{2}}}
$$

Squaring and rearranging,

$$
\frac{R^{2}}{m^{2}}=\frac{r^{2}}{m(m-2 r)}
$$

from which

$$
f(m)=R^{2} m=\frac{r^{2} m^{2}}{m-2 r}
$$

we need to find its minimum.

A differentiable function can have an extremum at the boundary of its domain, or where its derivative is 0. In our case, $m>2 r$, and as $m \rightarrow 2 r$ or $m \rightarrow \infty$, $f(m) \rightarrow+\infty$. Furthermore,

$$
f^{\prime}(m)=\frac{2 r^{2}(m-2 r)-r^{2} m^{2}}{(m-2 r)^{2}}=\frac{r^{2} m^{2}-4 r^{3} m}{(m-2 r)^{2}}
$$

Since the denominator is positive, the fraction is 0 when the numerator is 0, i.e.,

$$
r^{2} m^{2}-4 r^{3} m=r^{2} m(m-4 r)=0
$$

$r^{2} m \neq 0$, since $r$ and $m>0$, if $m-4 r=0$, then $m=4 r$.

It is also necessary to verify that there is indeed an extremum (and specifically a minimum) at this point, for example, by checking if the derivative changes sign (from negative to positive). Some have written that where the derivative is 0, the function has an extremum, which is not necessarily true.

If $m=4 r$, then the sought ratio, according to (1),

$$
\frac{V_{2}}{V_{1}}=\frac{r(4 r)^{2}}{4 r-2 r} \cdot \frac{1}{6 r^{3}}=\frac{4}{3}
$$

Based on the work of Éva Kovács (Vác, Friss I. Közg. Szki., III. o. t).

Note. The extremum can also be determined without differentiation if the variable is chosen wisely. This is demonstrated in the work of Tamás Bérczes, 8th grade, Vésztő, P. Szabó Ált. Isk.