A right circular cone is inscribed in a sphere of radius $r$. At what distance from the center of the sphere should a plane parallel to the base of the cone be placed so that the difference in the areas of the resulting sections is maximum or minimum?
Let $x$ be the distance of the intersecting plane from the center of the sphere; $y$ and $z$ the radii of the circles cut from the cone and the sphere, respectively. Since the radius of the base of the equilateral cone inscribed in the sphere is $\frac{r}{2} \sqrt{3}$, and its height is $\frac{3}{2} r$, we have:

$$
\frac{3}{2} r: r-x=\frac{r}{2} \sqrt{3}: y
$$

from which

$$
y=\frac{r-x}{\sqrt{3}}
$$

Furthermore,

$$
z^{2}=(r-x)(r+x)
$$

Therefore, the difference in the areas of the sections is

$$
\begin{aligned}
\left(z^{2}-y^{2}\right) \pi & =\pi\left[(r-x)(r+x)-\frac{\left(r-x^{2}\right.}{3}\right]= \\
& =\frac{\pi}{3}\left(-4 x^{2}+2 r x+2 r^{2}\right)
\end{aligned}
$$

This function attains its maximum value when

$$
x=\frac{r}{4}
$$

In this case, the difference in areas is $\frac{3}{4} \pi r^{2}$.

The difference in areas is the smallest when the intersecting plane passes through the base or the vertex of the cone. If, therefore, $x$ varies from $-\frac{r}{2}$ to $+r$, then the difference in areas increases from 0 to $\frac{3}{4} \pi r^{2}$, and then decreases again to 0.

The problem was also solved by: Burján K., Kohn B., Krausz B., Krisztián Gy., Lukhaub Gy., Lupsa Gy., Obláth R., Pollák L., Weisz J.