A parallelepiped has one vertex at $A$, the vertices adjacent to $A$ are $B, C$, and $D$, and the vertex opposite to $A$ is $E$. Draw a plane through $E$ that intersects the rays $A B, A C, A D$. Denote the intersection points as $P$, $Q$, and $R$.

How should we choose the plane so that the volume of the tetrahedron $A P Q R$ is minimized?
According to the formula found in the school function table, the volume $V$ of the tetrahedron $A P Q R$ is

$$
V=\frac{1}{6} \cdot A P \cdot A R \cdot A Q \cdot \sin (A P, A Q) \angle \cdot \sin \delta
$$

where $\delta$ is the angle of inclination between the plane $A P Q$ and the edge $A R$. Since $(A P, A Q) \angle$ and $\delta$ are given, $V$ is minimal if and only if $A P \cdot A R \cdot A Q$ is minimal.

Applying problem 1991 from the first volume of "Collection of Geometric Problems" to our case:

$$
\frac{A B}{A P}+\frac{A C}{A Q}+\frac{A D}{A R}=1
$$

Based on the relationship between the arithmetic and geometric means:

$$
\frac{1}{3}=\frac{\frac{A B}{A P}+\frac{A C}{A Q}+\frac{A D}{A R}}{3} \geq \sqrt[3]{\frac{A B \cdot A C \cdot A D}{A P \cdot A Q \cdot A R}}
$$

from which

$$
A P \cdot A Q \cdot A R \geq 27 A B \cdot A C \cdot A D
$$

(2) and (3) are equivalent, and the right-hand side of (3) is constant, so $A P \cdot A Q \cdot A R$, and thus $V$, are maximal if and only if equality holds in (2), that is, if

$$
\frac{A B}{A P}=\frac{A C}{A Q}=\frac{A C}{A R}=\frac{1}{3}
$$

since their sum is 1 according to (1). We note that (1) also ensures that the points $P, Q, R$, and $E$ lie in the same plane, corresponding to the relationships $A P=3 A B, A Q=3 A C, A R=3 A D$.