\section*{Task 2 - 071042}

For a body in the shape of a pyramid with a square base and congruent side faces, a cuboid container of the smallest possible volume is to be made. The pyramidal body is to be placed in such a way that it touches one of the inner faces of the container either with its base or with one of its side faces. Let \(h\) be the height of the pyramidal body and \(a\) the side length of its base.

Investigate which of these two positions requires a container with a smaller volume! It is useful to distinguish the cases \(h < \frac{a}{2}\) and \(h \geq \frac{a}{2}\).
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Let \(V_{1}\) be the volume of the container for the first of the above-mentioned layers of the pyramid. The base area of the pyramid then coincides with one side face of the enclosing cuboid, and the height of the pyramid is equal to the height of the container. It holds that

\[
V_{1}=a^{2} h
\]

Let \(V_{2}\) be the volume of the container in the second position of the pyramid. How do we calculate \(V_{2}\) for the two possible cases \(h \geq \frac{a}{2}\) and \(h < \frac{a}{2}\)?

For the case \(h \geq \frac{a}{2}\):

If the angle is \(90^{\circ}\), then \(s_{1}=M Q Y M S=s\). For the volume of the enclosing cuboid, it holds (with \(x=M^{\prime} Q\)):

\[
V_{2}=a s_{1} x
\]

As in Case 1, \(x=\frac{a h}{s}\); therefore,

\[
V_{2}=a^{2} h \cdot \frac{s_{1}}{s}
\]

and since \(\frac{s_{1}}{s}>1\), it follows that \(V_{2}>V_{1}\).

The volume is smaller when the base area of the pyramid coincides with one side face of the container.

\section*{Adapted from \([5]\)}