\section*{Problem 21 - V01221}

The cross-section of a sewer pipe is to have the shape of a rectangle with a semicircle on top.

What height and width should it be given if the area of the cross-section is \(1 \mathrm{~m}^{2}\) and the manufacturing costs are to be as low as possible?

It should be taken into account that the construction site allows a maximum height of \(0.9 \mathrm{~m}\).
Let \(h>0\) be the height measured in meters and \(b>0\) analogously the width of the rectangle. Then its cross-sectional area measured in square meters is \(1=h \cdot b+\frac{\pi}{8} \cdot b^{2}\) and thus \(h=b^{-1}-\frac{\pi}{8} \cdot b\). Furthermore, the channel has a height measured in meters of \(0.9 \geq h+\frac{b}{2}\).

The production costs depend monotonically on the perimeter of the cross-sectional area of the channel, so this and with it the term \(2 h+b+\frac{\pi}{2} \cdot b\) must be minimized under the mentioned constraints. If we substitute the relationship between \(h\) and \(b\) obtained from the size of the cross-sectional area, we get the term

\[
f(b)=2 b^{-1}-\frac{\pi}{4} \cdot b+b+\frac{\pi}{2} \cdot b=2 b^{-1}+\frac{\pi+4}{4} \cdot b
\]

which has the derivative \(f^{\prime}(b)=\frac{\pi+4}{4}-2 b^{-2}\), which vanishes exactly for \(b=\sqrt{\frac{\pi+4}{8}}\). A brief consideration for \(b \rightarrow 0\) and \(b \rightarrow \infty\) shows that \(f(b)\) goes to infinity in both cases, so it is monotonically decreasing up to the only critical point and then monotonically increasing; thus, there is a global minimum at the critical point.

The production costs are therefore - without considering the height constraint - minimal when \(b=\sqrt{\frac{\pi+4}{8}} \approx\) \(0.945 \, \text{m}\) and \(h=b^{-1}-\frac{\pi}{8} \cdot b \approx 0.687 \, \text{m}\). However, the channel would then have a total height of more than \(0.9 \, \text{m}\). Therefore, the width \(b\) of the channel must be adjusted so that this height constraint is met.

Since any increase in \(b\) beyond the critical value or any decrease below it further increases the perimeter of the cross-sectional area and thus the production costs, they will be minimal under consideration of the height constraint when the equality case holds in the height constraint.

Thus, we can now additionally assume \(0.9=h+\frac{b}{2}\). Substituting this into the relationship between \(h\) and \(b\) obtained from the cross-sectional area, we get the equation

\[
0.9-\frac{b}{2}=h=b^{-1}-\frac{\pi}{8} \cdot b \quad \text{or} \quad \frac{4-\pi}{8} \cdot b^{2}-0.9 b+1=0
\]

which has the two solutions

\[
b_{1}=\frac{3.6}{4-\pi}+\sqrt{\frac{3.6^{2}}{(4-\pi)^{2}}-\frac{8}{4-\pi}} \approx 7 \quad \text{and} \quad b_{2}=\frac{3.6}{4-\pi}-\sqrt{\frac{3.6^{2}}{(4-\pi)^{2}}-\frac{8}{4-\pi}} \approx 1.318
\]

The first solution is discarded because only a negative \(h\) could then satisfy the height constraint, which is excluded. Therefore, \(b=b_{2}\) must hold, and we get \(h=0.9-\frac{b}{2} \approx 0.241\).

Thus, the rectangle underlying the channel must have a width of approximately \(1.318 \, \text{m}\) and a height of approximately \(0.241 \, \text{m}\).