Among the cylinders that can be inscribed in a sphere, determine the one whose total surface area is the largest possible.
If the radius of a sphere is $R$, the radius of the base of a cylinder is $r$, and its height is $m$, then the total surface area of the cylinder is:

$$
F=2 r \pi(r+m)
$$

but

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

so

$$
F=2 r \pi\left(r+2 \sqrt{R^{2}-r^{2}}\right)
$$

or

$$
\begin{gathered}
\frac{F}{2 r \pi}-r=2 \sqrt{R^{2}-r^{2}} \\
\frac{F^{2}}{4 r^{2} \pi^{2}}-\frac{F}{\pi}+r^{2}=4 R^{2}-4 r^{2}
\end{gathered}
$$

rearranging:

$$
r^{4}-\frac{2}{5}\left(\frac{F}{2 \pi}+2 R^{2}\right) r^{2}=-\frac{F^{2}}{20 \pi^{2}}
$$

from which

$$
r=\sqrt{\frac{1}{5}\left(\frac{F}{2 \pi}+2 R^{2}\right) \pm \sqrt{\frac{1}{25}\left(\frac{F}{2 \pi}+2 R^{2}\right)^{2}-\frac{F^{2}}{20 \pi^{2}}}}
$$

$r$ has a real value if

$$
\frac{1}{5}\left(\frac{F}{2 \pi}+2 R^{2}\right)^{2} \geqq \frac{F^{2}}{4 \pi^{2}}
$$

or if

$$
F \leqq R^{2} \pi(1+\sqrt{5})
$$

and thus the maximum surface area is:

$$
F=R^{2} \pi(1+\sqrt{5})
$$

Substituting this value of $F$ into (2) and (1), we get:

$$
r=R \sqrt{\frac{5+\sqrt{5}}{10}} \text { and } m=2 R \sqrt{\frac{5-\sqrt{5}}{10}}
$$

(Jenő Filkorn, Nyitra.)

The problem was also solved by: Faith F., Kerekes T., König D., Krausz B., Gy. Krisztián, Gy. Lukhaub, Gy. Lupsa, J. Scharff.