A homeowner wants to have a water tank built. This water tank must comply with the following specifications:
\begin{itemize}
  \item it must be located two metres from his house;
  \item the maximum depth must be two metres;
  \item it must measure five metres long;
  \item it must follow the natural slope of the land.
\end{itemize}

The curved part is modelled by the curve $\mathscr{C}_f$ of the function $f$ on the interval $[2; 2e]$ defined by:
$$f(x) = x \ln\left(\frac{x}{2}\right) - x + 2$$

The curve $\mathscr{C}_f$ is represented in an orthonormal coordinate system with unit $1\mathrm{m}$ and constitutes a profile view of the tank.\\
We consider the points $\mathrm{A}(2; 2)$, $\mathrm{I}(2; 0)$ and $\mathrm{B}(2\mathrm{e}; 2)$.

\textbf{Part A}

The objective of this part is to evaluate the volume of the tank.

\begin{enumerate}
  \item Justify that the points B and I belong to the curve $\mathscr{C}_f$ and that the x-axis is tangent to the curve $\mathscr{C}_f$ at point I.
  \item We denote by $\mathscr{T}$ the tangent to the curve $\mathscr{C}_f$ at point B, and D the point of intersection of the line $\mathscr{T}$ with the x-axis.\\
a. Determine an equation of the line $\mathscr{T}$ and deduce the coordinates of D.\\
b. We call $S$ the area of the region bounded by the curve $\mathscr{C}_f$, the lines with equations $y = 2$, $x = 2$ and $x = 2\mathrm{e}$.\\
$S$ can be bounded by the area of triangle ABI and that of trapezoid AIDB.\\
What bounds on the volume of the tank can we deduce?
  \item a. Show that, on the interval $[2; 2\mathrm{e}]$, the function $G$ defined by
$$G(x) = \frac{x^2}{2} \ln\left(\frac{x}{2}\right) - \frac{x^2}{4}$$
is an antiderivative of the function $g$ defined by $g(x) = x \ln\left(\frac{x}{2}\right)$.\\
b. Deduce an antiderivative $F$ of the function $f$ on the interval $[2; 2\mathrm{e}]$.\\
c. Determine the exact value of the area $S$ and deduce an approximate value of the volume $V$ of the tank to the nearest $\mathrm{m}^3$.
\end{enumerate}

\textbf{Part B}

For any real number $x$ between 2 and $2\mathrm{e}$, we denote by $v(x)$ the volume of water, expressed in $\mathrm{m}^3$, in the tank when the water level in the tank is equal to $f(x)$.\\
We admit that, for any real number $x$ in the interval $[2; 2\mathrm{e}]$,
$$v(x) = 5\left[\frac{x^2}{2}\ln\left(\frac{x}{2}\right) - 2x\ln\left(\frac{x}{2}\right) - \frac{x^2}{4} + 2x - 3\right]$$

\begin{enumerate}
  \item What volume of water, to the nearest $\mathrm{m}^3$, is in the tank when the water level in the tank is one metre?
  \item We recall that $V$ is the total volume of the tank, $f$ is the function defined at the beginning of the exercise and $v$ the function defined in Part B.\\
We consider the following algorithm:\\
\begin{tabular}{|l|l|}
\hline
\begin{tabular}{l}
Variables: \\
Processing: \\
\end{tabular} & \begin{tabular}{l}
$a$ is a real number \\
$b$ is a real number \\
$a$ takes the value 2 \\
$b$ takes the value $2\mathrm{e}$ \\
While $v(b) - v(a) > 10^{-3}$ do: \\
$c$ takes the value $(a + b)/2$ \\
If $v(c) < V/2$, then: \\
$a$ takes the value $c$ \\
Otherwise \\
$b$ takes the value $c$ \\
End If \\
End While \\
Display $f(c)$ \\
\end{tabular} \\
\hline
\end{tabular}\\
Interpret the result that this algorithm allows to display.
\end{enumerate}