\section*{Problem 18 - V01118}

A straight circular cylinder is to be inscribed in a given sphere.

What should be the ratio of the height \(h\) to the diameter \(d\) of the cylinder so that

a) the volume,

b) the lateral surface area,

c) the total surface area of the cylinder is maximized?
Let \(R\) be the radius of the sphere, \(r\) the radius of the cylinder, and \(h\) the height of the cylinder. To find their relationship, we place a cutting plane through the sphere and cylinder, perpendicular to the top and bottom faces of the cylinder and passing through the center of the sphere.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1666.jpg?height=340&width=371&top_left_y=2120&top_left_x=291)

In the figure, the axis of rotation of the cylinder is shown in red. For the right triangle, the Pythagorean theorem gives

\[
R^{2}=r^{2}+\left(\frac{h}{2}\right)^{2}
\]

a) The volume of the cylinder is given by

\[
V(r, h)=\pi r^{2} h
\]

Rearranging (1) for \(r^{2}\) and substituting, we get

\[
V(h)=\pi\left(R^{2}-\frac{h^{2}}{4}\right) h
\]

The roots of the first derivative \(V^{\prime}(h)=\pi\left(R^{2}-\frac{3 h^{2}}{4}\right)\) are

\[
h_{1,2}= \pm \frac{2 \sqrt{3}}{3} R
\]

The negative solution is discarded. The second derivative \(V^{\prime \prime}=-\frac{3}{2} \pi h\) is negative for all positive \(h\). Therefore, a local maximum exists.

The cylinder has a maximum volume with the radius \(r=\frac{2 \sqrt{3}}{3} R\) and the height \(h=\frac{4 \sqrt{3}}{3} R\). The ratio is \(h: r=\sqrt{2}: 1\).

b) The lateral surface area of the cylinder is given by

\[
M=2 \pi r \cdot h
\]

Substituting (1) again, we get the objective function

\[
M=2 \pi \sqrt{R^{2}-\frac{h^{2}}{4}} h
\]

with the first derivative

\[
M^{\prime}=\pi \sqrt{4 R^{2}-h^{2}}-\frac{\pi h^{2}}{\sqrt{4 R^{2}-h^{2}}}
\]

The roots are obtained by some rearrangement as \(h_{1,2}=\sqrt{2} R\). \(h=\sqrt{2} R\) is again found to be the local maximum with some computational effort.

The cylinder has a maximum lateral surface area with the radius \(r=\frac{\sqrt{2}}{2} R\) and the height \(h=\sqrt{2} R\). The ratio is \(h: r=2: 1\).

c) The total surface area of the cylinder is given by

\[
O=2 \pi r \cdot h+2 \pi r^{2}
\]

Substituting (1) again, we get the objective function

\[
O=\pi h \sqrt{4 R^{2}-h^{2}}-\frac{\pi\left(h^{2}-4 R^{2}\right)}{2}
\]

with the first derivative

\[
O^{\prime}=\pi \sqrt{4 R^{2}-h^{2}}-\frac{\pi h^{2}}{\sqrt{4 R^{2}-h^{2}}}-\pi h
\]

A root that also gives the local maximum is \(h=\sqrt{\frac{2 \sqrt{5}}{5}+2} R\). (The computational proof is very involved). The cylinder has a maximum total surface area with the radius \(r=\sqrt{\frac{1}{2}-\frac{\sqrt{5}}{10}} R\) and the height \(\sqrt{\frac{2 \sqrt{5}}{5}+2} R\). The ratio is \(h: r=(1+\sqrt{5}): 1\).