616. An irrigation channel has the shape of an isosceles trapezoid, the lateral sides of which are equal to the smaller base. At what angle of inclination of the lateral sides is the cross-sectional area of the channel the largest?
Solution. Let the smaller base of the trapezoid be denoted by $a$, the angle of inclination of the lateral sides by $\alpha$, and the area of the section by $S$ (Fig. 124). According to the condition, $|A B|=|A D|=|B C|=a$.

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-261.jpg?height=215&width=356&top_left_y=1598&top_left_x=122)

Fig. 124 From triangle $B C K$, we find $|B K| = a \cos \alpha$. Then $|C D| = a + 2a \cos \alpha$. The height of the trapezoid is $|C K| = a \sin \alpha$. Let's determine the area of the trapezoid:

$$
\begin{gathered}
S = \frac{|A B| + |D C|}{2} |C K| = \frac{a + a + 2a \cos \alpha}{2} a \sin \alpha = \\
= a^2 (1 + \cos \alpha) \sin \alpha.
\end{gathered}
$$

To find the maximum value of the area $S = S(\alpha)$, we need to find the critical points within the interval $(0, \pi / 2)$, compute the values of the function $S(\alpha)$ at these points and at the endpoints of the interval $(0, \pi / 2)$, and then select the largest value from the obtained results.

Let's find the derivative:

$$
\begin{aligned}
& S'(\alpha) = a^2 \left( (1 + \cos \alpha)' \sin \alpha + (1 + \cos \alpha) (\sin \alpha)' \right) = \\
= & a^2 \left( -\sin^2 \alpha + \cos \alpha + \cos^2 \alpha \right) = a^2 \left( 2 \cos^2 \alpha + \cos \alpha - 1 \right)
\end{aligned}
$$

Set the derivative to zero and solve the resulting equation:

$$
2 \cos^2 \alpha + \cos \alpha - 1 = 0 ; \quad \cos \alpha_1 = 1 / 2 ; \quad \cos \alpha_2 = -1
$$

from which $\alpha_1 = \pm \pi / 3 + 2 \pi n, \alpha_2 = \pi + 2 \pi n$. Only the critical point $\alpha_1 = \pi / 3$ belongs to the interval $(0, \pi / 2)$.

Let's find $S(\alpha_1) = S\left(\frac{\pi}{3}\right) = a^2 \left(1 + \cos \frac{\pi}{3}\right) \sin \frac{\pi}{3} = \frac{3 \sqrt{3}}{4} a^2, S(0) = 0$, $S\left(\frac{\pi}{2}\right) = a^2$. Comparing the obtained results, we conclude that the function $S(\alpha)$ attains its maximum value at $\alpha = \pi / 3$.

Thus, the area of the channel section is the largest when the angle of inclination of the lateral sides is $60^{\circ}$.