Let's inscribe a frustum of a cone with maximum surface area in a given hemisphere. (One of the base circles of the frustum should be the great circle bounding the hemisphere.)
The cross-section of the frustum is a trapezoid inscribed in a semicircle cut from a hemisphere. The longer parallel side of the trapezoid is the diameter of the semicircle. We choose this as our unit of length and express the slant height $a$ and the smaller base diameter $d$ of the frustum - the leg and the shorter parallel side of the trapezoid - in terms of the angle $\varphi$ between the diagonal and the parallel sides:

$$
a=\sin \varphi=t, \quad d=1-2 \sin ^{2} \varphi=1-2 t^{2}, \quad \text { where } t=\sin \varphi
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_94fbdb13f95319eeee88g-1.jpg?height=301&width=526&top_left_y=451&top_left_x=794)

With these, the variable part of the frustum's surface area (omitting the constant area of the lower base circle) is

$$
F=\left(\frac{d}{2}\right)^{2} \pi+\frac{\pi+d \pi}{2} \cdot a
$$

and it will be useful to examine the following function:

$$
y=\frac{F}{\pi}-\frac{1}{4}=\left(\frac{1}{2}-t^{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+\left(1-t^{2}\right) t=t^{4}-t^{3}-t^{2}+t
$$

where

$$
00$, the function increases, and in the latter, $y^{\prime}<0$, the function decreases, thus proving our statement.

At the found maximum point, the slant height and the diameter of the top circle of the frustum are

$$
a=t_{0}=\frac{\sqrt{17}-1}{8}=0.390, \quad d=1-2 t_{0}^{2}=\frac{7+\sqrt{17}}{16}=0.696
$$

and the total surface area (including the base circle of the hemisphere) is

$$
\frac{\pi}{512} \cdot(149+51 \sqrt{17})=\pi \cdot 0.702=2.20 \quad \text { square units. }
$$

Peter Gál (Budapest, Fazekas M. Gymnasium, 2nd year)