\section*{Problem 17 - V0117}

A drainage channel has as its inner cross-section a rectangle with a semicircle placed on top. What dimensions must the channel have so that the cross-section is as large as possible when the perimeter \(U\) is constant?

How large is the largest cross-section?
Let \(a\) be the width of the channel, \(b\) the height of the rectangular part, and \(r=\frac{a}{2}\) the radius of the attached semicircle. For the perimeter of the channel, we then have

\[
u=a+2 b+\pi a \quad \rightarrow \quad b=\frac{u-a-\pi a}{2}
\]

The cross-sectional area consists of the rectangle and the semicircle:

\[
A=a \cdot b+\frac{1}{2} \pi\left(\frac{a}{2}\right)^{2}
\]

Substituting (1) into (2) and simplifying yields the objective function for the area:

\[
A(a)=\frac{a \cdot u}{2}-\frac{a^{2}(3 \pi+4)}{8}
\]

with the derivatives

\[
A^{\prime}(a)=\frac{u}{2}-\frac{a(3 \pi+4)}{4} \quad ; \quad A^{\prime \prime}(a)=-\frac{3 \pi+4}{4}<0
\]

The zero of the first derivative is \(a=\frac{2 u}{3 \pi+4}\). Since the second derivative is always negative, there is a local maximum.

For \(b\), we get \(b=\frac{u(\pi+2)}{6 \pi+8}\).

The channel should have a width of \(a=\frac{2 u}{3 \pi+4}\) and a rectangular height of \(b=\frac{u(\pi+2)}{6 \pi+8}\) to achieve a maximum cross-sectional area of \(A=\frac{u^{2}}{6 \pi+8}\).