Let's inscribe a cone in a sphere, such that the volume of the cone is maximum.
Let the radius of the sphere be $R$, the radius of the base of the cone be $x$, and its height be $y$.

The volume of the cone is:

$$
K=\frac{x^{2} \cdot y \pi}{3}
$$

but since

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

therefore

$$
K=\frac{\pi}{3} \cdot y^{2}(2 R-y)
$$

The volume of the cone will be maximum when the function

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

takes its maximum value.

Multiplying both sides of the equation by 2:

$$
2 z=y \cdot y(4 R-2 y)
$$

However, the sum of the factors on the right side of the equation $(4 R)$ is a constant number, which is why the volume of the cone will be the largest possible when the factors are equal (K.M.L.V.36.1.), that is, when

$$
4 R-2 y=y
$$

from which the height of the cone is:

$$
y=\frac{4}{3} R
$$

in which case the radius of the base of the cone is

$$
x=\frac{2}{3} R \sqrt{2}
$$

(Krisztián György, Pécs.)

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