Among the right circular cylinders that can be inscribed in a sphere, for which cylinder will the ratio of the surface area to the volume be the smallest?
I. solution. Let the radius of the sphere be $R$, the radius of the base circle of the cylinder be $r$, and the height of the cylinder be $h$. Since the center of the cylinder and the sphere coincide, by the Pythagorean theorem, $h=2 \sqrt{R^{2}-r^{2}}$, and the ratio of the surface area to the volume is

$$
\frac{A}{V}=\frac{2 \pi r(h+r)}{\pi r^{2} h}=\frac{2}{r}+\frac{2}{h}+\frac{2}{r}=\frac{1}{\sqrt{R^{2}-r^{2}}}
$$

Apply the weighted harmonic and quadratic mean inequality to the numbers $\frac{r}{\sqrt[3]{2}}$ and $\sqrt{R^{2}-r^{2}}$ with weights $\sqrt[3]{4}$ and 1:

$$
(\sqrt[3]{4}+1) \frac{V}{A}=\frac{\sqrt[3]{4}+1}{\sqrt[3]{4} \cdot \frac{\sqrt[3]{2}}{r}+\frac{1}{\sqrt{R^{2}-r^{2}}}} \leq \sqrt{\frac{\sqrt[4]{4} \cdot\left(\frac{r}{\sqrt[3]{2}}\right)^{2}+{\sqrt{R^{2}-r^{2}}}^{2}}{\sqrt[3]{4}+1}}=\frac{R}{\sqrt{\sqrt[3]{4}+1}}
$$

Rearranging,

$$
\frac{A}{V} \geq \frac{(\sqrt[3]{4}+1)^{\frac{3}{2}}}{R}
$$

We have managed to estimate the ratio of the surface area to the volume from below with a quantity that depends only on the radius of the sphere. Equality holds if and only if the same two numbers were used for the means, i.e., $\frac{r}{\sqrt[3]{2}}=\sqrt{R^{2}-r^{2}}$. Solving for $r$,

$$
r=\frac{\sqrt[3]{2}}{\sqrt{\sqrt[3]{4}+1}} R \quad \text { and } \quad h=2 \sqrt{R^{2}-r^{2}}=\frac{2}{\sqrt{\sqrt[3]{4}+1}} R
$$

In this case, the ratio of the surface area to the volume is minimal, i.e., $\frac{(\sqrt[3]{4}+1)^{\frac{3}{2}}}{R}$.

II. solution. Let $f(x)=\frac{2}{x}+\frac{1}{\sqrt{R^{2}-x^{2}}}$. According to the previous solution, we need to find the minimum of this function in the interval $(0 ; R)$. We do this by examining the sign of the derivative.

$$
f^{\prime}(x)=-\frac{2}{x^{2}}+\frac{x}{\left(R^{2}-x^{2}\right)^{\frac{3}{2}}}=\frac{x^{3}-2\left(R^{2}-x^{2}\right)^{3 / 2}}{x^{2}\left(R^{2}-x^{2}\right)^{3 / 2}}
$$

In the interval under consideration, the denominator of the last fraction is always positive, and the numerator is strictly increasing. Its value is 0 when $x^{3}-2\left(R^{2}-x^{2}\right)^{3 / 2}=0$, i.e., $x=\frac{\sqrt[3]{2}}{\sqrt{\sqrt[3]{4}+1}} R$.

In the interval $\left(0 ; \frac{\sqrt[3]{2}}{\sqrt{\sqrt[3]{4}+1}} R\right)$, $f^{\prime}<0$, and $f$ is strictly decreasing. In the interval $\left(\frac{\sqrt[3]{2}}{\sqrt{\sqrt[3]{4}+1}} R ; R\right)$, $f^{\prime}>0$, and $f$ is strictly increasing. Therefore, the minimum of the function is at the point $x=\frac{\sqrt[3]{2}}{\sqrt{\sqrt[3]{4}+1}} R$.

Remark. Among those who solved the problem using derivatives, a common error was that they only looked for the zero of the derivative but did not, or incorrectly justified, why there is a minimum there. It is also not a complete solution if someone only determined that the function has a local minimum at the given point based on the sign of the second derivative.