We make a channel from three boards of equal width. The side walls form an angle $x$ with the base board. At what value of $x$ will the cross-section of the channel (and thus its volume) be the largest possible?

The side walls form an angle $x$ with the base board. At what value of $x$ will the cross-section of the channel (which is in the shape of an isosceles trapezoid) be the largest possible?
I. Solution: Let $x$ be the acute angle between the side wall and the baseboard. The further notations are shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_32b762a1db1f15766410g-1.jpg?height=363&width=599&top_left_y=266&top_left_x=752)

From the right-angled triangle $A F D$, the height of the trapezoid is:

$$
m=\sqrt{a^{2}-l^{2}}
$$

Thus, the area of the trapezoid is

$$
t=\frac{a+a+2 l}{2} m=(a+l) \sqrt{a^{2}-l^{2}}
$$

Since the area is always positive, $t$ (as a function of $l$) has its maximum where $3 t^{2}$ does:

$$
3 t^{2}=3(a+l)^{2}\left(a^{2}-l^{2}\right)=(a+l)(a+l)(a+l)(3 a-3 l)
$$

The sum of the four factors is $6 a$, a constant. Therefore, based on the arithmetic-geometric mean inequality, the product is maximal when the factors are equal:

$$
a+l=3 a-3 l
$$

which means

$$
l=\frac{a}{2}
$$

This determines the maximum cross-sectional area of the channel. Let's calculate the angle $x$ when the trapezoid has the maximum area. From the right-angled triangle $A F D$:

$$
\cos x=\frac{l}{a}
$$

In the case of the maximum area trapezoid, this value is $\frac{\frac{a}{2}}{a}=\frac{1}{2}$, so

$$
x=60^{\circ}
$$

Thus, the volume of the channel is maximized when the side wall forms a $60^{\circ}$ acute angle with the baseboard.

Gabor Pal (Miskolc, Kohaszati techn. II. o. t.)

Remarks: 1. Essentially, we arrive at the same solution if we directly express the cross-sectional area in terms of the angle $x$ between the base and the side wall. In this case,

$$
\begin{aligned}
& \quad l=a \cos x, \quad m=a \sin x \\
& \text { thus } \\
& t=\frac{2 a+2 l}{2} m=a^{2}(1+\cos x) \sin x \\
& \text { This has its maximum at the same point as } \\
& \left(\frac{t}{a^{2}}\right)^{2}=(1+\cos x)^{2} \sin ^{2} x=(1+\cos x)^{2}\left(1-\cos ^{2} x\right)= \\
& =(1+\cos x)(1+\cos x)(1+\cos x)(1-\cos x)
\end{aligned}
$$

and from this, multiplying by 3, the maximum can be determined similarly to the previous solution.

2. The problem can also be solved using differential calculus. We accepted solutions using this method as well.