The circumference of the cross-section of a cylinder is $90 \mathrm{~cm}$. What is the maximum possible volume of the cylinder?
Solution. The cross-section of the cylinder along its axis is a rectangle, one side of which is the diameter $d$ of the cylinder's base, and the other side is the height $m$ of the cylinder. We know that the perimeter of this rectangle is $90 \mathrm{~cm}$: $2(d+m)=90$. The volume of the cylinder is: $V=r^{2} \pi \cdot m$. We rearrange the perimeter formula:

$$
d+m=45 \quad \Longrightarrow \quad r+r+m=45
$$

For these three positive numbers, according to the inequality between the arithmetic mean and the geometric mean:

$$
\frac{r+r+m}{3} \geq \sqrt[3]{r \cdot r \cdot m}
$$

The value on the left side is $\frac{45}{3}=15$; the right side will be the largest when equality holds, and the necessary and sufficient condition for this is $r=r=m$.

If $\sqrt[3]{r \cdot r \cdot m}$ is the largest possible, then $r \cdot r \cdot m$ is also the largest possible, and $r^{2} \cdot m \cdot \pi$ (i.e., the volume) is also the largest possible. Therefore, $r=r=m=\frac{45}{3}=15 \mathrm{~cm}, V_{\max }=r^{2} \pi \cdot m=15^{3} \cdot \pi$, which means the volume of the cylinder can be at most $15^{3} \cdot \pi \approx 10603 \mathrm{~cm}^{3}$.