We want to build a cement storage tank in the shape of a prism. The base is formed from a square with a side length of $2 \mathrm{~m}$, such that two of the base's additional vertices fall on one midline of the square, and the other two fall on the other midline. The height of the prism is 2 $\mathrm{m}$. The foundation, floor, and walls cost $1000 \mathrm{Ft}$ per $\mathrm{m}^{2}$, and the completed storage tank costs $2500 \mathrm{Ft}$ per $\mathrm{m}^{3}$. How should we shape the base to maximize the benefit of the construction?
Let the starting base square be $ABCD = N$, the entire base of the domain be two consecutive sides $AA_1$ and $A_1B$, and the center of $N$ be $O$. First, let's examine the part of the to-be-constructed storage $T$ over the angle domain $AOB$, denoted by $T_1$.
![](https://cdn.mathpix.com/cropped/2024_05_02_83e729a81c29c43e3749g-1.jpg?height=628&width=1202&top_left_y=298&top_left_x=450)

Let $\angle OAA_1 = x$, then (since $OA = \sqrt{2}$),

\[
\begin{gathered}
AA_1 = \frac{OA \sin 45^\circ}{\sin x} = \frac{1}{\sin x} \\
OA_1 = OF + FA_1 = 1 + \cot x
\end{gathered}
\]

where $F$ is the midpoint of side $AB$. Therefore, the volume of $T_1$ is

\[
V_1 = 2 \frac{OA_1 \cdot AB}{2} = 2(1 + \cot x)
\]

and the total surface area of the bounding faces of $T_1$ is

\[
F_1 = 2 \cdot \frac{OA_1 \cdot AB}{2} + 2(AA_1 + A_1B) = 2(1 + \cot x) + \frac{4}{\sin x}
\]

The profit from constructing the part $T_1$ of the storage (choosing the unit as a thousand forints):

\[
H_1 = H_1(x) = 2.5 V_1 - F_1 = 3(1 + \cot x) - \frac{4}{\sin x}
\]

According to the problem statement, this function is $H_1'(x) < 0$ if $x < 41^\circ 25'$, and $H_1'(x) > 0$ if $x > 41^\circ 25'$, so $H_1(x)$ has a maximum at $x_0 = 41^\circ 25'$. (In this case, point $A_1$ is outside $N$.)

A similar result is obtained by examining the part of the storage $T$ over the angle domains $BOC$, $COD$, and $DOA$; the profit from constructing these parts depends only on the angles $\angle OBB_1$, $\angle OCC_1$, and $\angle DDO_1$, and is maximized independently when $\angle OBB_1 = \angle OCC_1 = \angle DDO_1 = 41^\circ 25'$. In this case, $OA_1 = OB_1 = OC_1 = OD_1 = 2.13 \, \text{m}$, and the total profit from the construction is $4 H_1(x_0) = 1416 \, \text{Ft}$.

Lukács Éva (Debrecen, Fazekas M. Gym., II. o. t.) Katona Klára (Budapest, Fazekas M. Practical Gym., I. o. t.)