The problem describes a situation where a farmer, Madame BASSOA, provides metal sheets to a blacksmith to create cylindrical troughs. The goal is to determine the volume of the trough proposed by the blacksmith as a function of the width $l$ and length $L$ of the metal sheets. Also, explain how the blacksmith can obtain the value of $x$ given that point $O$ is not supported. We are given that the blacksmith bends the width $l$ into an arc of a circle, forming a trough. The angle subtended by the arc at the center $O$ is $x$. The length of the trough is $L$.

1. Problem Description

The problem describes a situation where a farmer, Madame BASSOA, provides metal sheets to a blacksmith to create cylindrical troughs. The goal is to determine the volume of the trough proposed by the blacksmith as a function of the width $l$ and length $L$ of the metal sheets. Also, explain how the blacksmith can obtain the value of $x$ given that point $O$ is not supported.

We are given that the blacksmith bends the width $l$ into an arc of a circle, forming a trough. The angle subtended by the arc at the center $O$ is $x$. The length of the trough is $L$.

2. Solution Steps

First, we need to find the volume of the trough as a function of $l$, $L$, and $x$.

The area of the segment of the circle is given by $S(x) = \frac{l^2}{2} \frac{(x - \sin x)}{x^2}$.

The radius $r$ of the circle satisfies $rx = l$, so $r = \frac{l}{x}$. The area of the segment can be calculated as the area of the sector minus the area of the triangle. The area of the sector is $\frac{1}{2} r^2 x = \frac{1}{2} (\frac{l}{x})^2 x = \frac{l^2}{2x}$. The area of the triangle is $\frac{1}{2} r^2 \sin x = \frac{1}{2} (\frac{l}{x})^2 \sin x = \frac{l^2 \sin x}{2x^2}$. Therefore the area of the segment is:

$A = \frac{l^2}{2x} - \frac{l^2 \sin x}{2x^2} = \frac{l^2}{2} (\frac{x - \sin x}{x^2})$.

The volume of the trough is the area of this segment multiplied by the length $L$. Therefore,

$V(x) = L \times A = L \frac{l^2}{2} (\frac{x - \sin x}{x^2}) = \frac{L l^2}{2} \frac{(x - \sin x)}{x^2}$.

To explain how the blacksmith can obtain the value of $x$, we need to consider the stability of the trough. Since point $O$ is not supported, the trough will be stable only if the center of mass of the cross-sectional area (the segment) lies directly below the center $O$. We're given that $\forall x \in \mathbb{R}_+$, $|\frac{x-\sin x}{x^2}| \le \frac{1}{6}|x|$, we can calculate values of $u(x) = \tan(x/2) - \frac{x}{2}$.

Because the bottom of the trough is an arc, the blacksmith will want to chose the $x$ value where the trough is stable. That is to choose a value where it will not flip to it's side, such as $x=\pi$.

3. Final Answer

The volume of the trough is $V(x) = \frac{L l^2}{2} \frac{(x - \sin x)}{x^2}$. The blacksmith needs to choose the value of $x$ such that the trough is stable.