Adott egyenes kúpba írjunk hengert, melynek köbtartalma maximum.

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Place a cylinder inside a given straight cone, such that the cylinder's volume is maximized.
Let the height of the cone be $m$, the radius of its base $R$, the height of the cylinder $x$, and the radius of its base $r$.

The volume of the cylinder is:

$$
K=\pi r^{2} \cdot x
$$

but from the similar triangles involved:

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

and thus

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

Therefore, the volume of the cylinder is maximized when the function

$$
y=r^{2}(R-r)
$$

takes its maximum value.

Multiplying both sides of the equation by 2:

$$
2 y=r \cdot r(2 R-2 r)
$$

The sum of the factors on the right side of the equation $(2 R)$ is a constant number, which is why the volume of the cylinder is maximized when the factors are equal, that is, when

$$
2 R-2 r=r
$$

from which

$$
r=\frac{2}{3} R
$$

in which case the height of the cylinder is $x=\frac{1}{3} m$.

(Krisztián György, Pécs.)

The problem was also solved by: Krausz B., Lukhaub Gy., Sasvári G.