A steam boiler consists of a cylinder that ends in a hemisphere at each end. If the length and width of the cylinder are changed so that the total length of the boiler remains constant, the question is, what dimensions of the cylinder will maximize the volume of the boiler?
Let $l$ be the length of the boiler, $x$ and $y$ be the diameter and height of the cylinder, respectively. The volume of the boiler is

$$
V=\pi\left(\frac{x}{2}\right)^{2} y+\frac{1}{6} \pi x^{3}
$$

Since $y=l-x$,

$$
V=\frac{\pi x^{2}}{12}(3 l-x)
$$

To find the maximum of $V$, it is sufficient to find the maximum of the expression

$$
x^{2}(3 l-x)
$$

This expression can be written as

$$
x^{p} y^{l}
$$

where $y=3 l-x, p=2, q=1$, and $x+y=3 l$.

The maximum of the expression $x^{p} y^{q}$, given that $x+y=3 l=$ constant, is obtained when

$$
\frac{x}{p}=\frac{y^{\star}}{q}
$$

In this case, it is when

$$
\frac{x}{2}=\frac{3 l-x}{1}
$$

which means $x=2 l$.

Thus, as $x$ increases from 0 to $2 l$, $V$ increases from 0 to $\frac{\pi l^{3}}{3}$ and reaches its maximum there. However, the maximum value of $x$ in the context of the problem is at most $l$, so $V$ never reaches its absolute maximum; nevertheless, the value $V=\frac{\pi l^{2}}{6}$ is the relatively largest value that can be achieved under the given conditions. In this case, the boiler will be spherical.

*) See Theorem III on page 44 of the first volume of "Középiskolai Math. Lapok".