In a cardboard disk of radius $R$, we cut out an angular sector corresponding to an angle of measure $\alpha$ radians. We overlap the edges to create a cone of revolution. We wish to choose the angle $\alpha$ to obtain a cone of maximum volume.

We call $\ell$ the radius of the circular base of this cone and $h$ its height.\\
We recall that:
\begin{itemize}
  \item the volume of a cone of revolution with base a disk of area $\mathscr{A}$ and height $h$ is $\frac{1}{3}\mathscr{A}h$.
  \item the length of an arc of a circle of radius $r$ and angle $\theta$, expressed in radians, is $r\theta$.
\end{itemize}

\begin{enumerate}
  \item We choose $R = 20\mathrm{~cm}$.\\
a. Show that the volume of the cone, as a function of its height $h$, is
$$V(h) = \frac{1}{3}\pi\left(400 - h^2\right)h.$$
b. Justify that there exists a value of $h$ that makes the volume of the cone maximum. Give this value.\\
c. How should we cut the cardboard disk to have maximum volume? Give an approximation of $\alpha$ to the nearest degree.
  \item Does the angle $\alpha$ depend on the radius $R$ of the cardboard disk?
\end{enumerate}