From a right circular cone with base radius $r=8 \mathrm{~cm}$ and height $m=15 \mathrm{~cm}$, we cut out a spherical cap such that its base circle lies on the base of the cone and the corresponding spherical segment touches the lateral surface of the cone. How does the volume of this spherical cap change? (The height of the spherical cap is the independent variable.)
We must examine those spheres whose centers lie on the axis of the cone and which touch the cone's surface, more precisely the cone surface defined by the data (also outside the body).

![](https://cdn.mathpix.com/cropped/2024_05_02_8a969ea9e6aa6b332c30g-1.jpg?height=545&width=550&top_left_y=259&top_left_x=777)

These spheres, in terms of position and size, vary between two limits. One limiting sphere is the sphere that touches the cone surface and the base of the cone from the outside, \( G_{1} \). The height of the spherical segment in this case is \( x=0 \), and its volume is \( y=0 \). The other limiting sphere is the sphere inscribed in the cone, \( G_{2} \). In this case, the spherical segment is the entire sphere, its height is the diameter of \( G_{2} \): \( x=2 \varrho_{2} \), and its volume is \( y=\frac{4 \varrho_{2}^{3} \pi}{3} \).

For any arbitrary sphere \( G \) between these two limits, let its radius be \( \varrho \), its center \( O \), the height of the spherical segment inside the cone \( x=D E \), and its volume

\[
y=\frac{x^{2} \pi}{3}(3 \varrho-x)
\]

We already express \( \varrho \) as a function of \( x \). According to our diagram,

\[
\begin{gathered}
A F O \Delta \sim A B D \Delta \text { and thus } \\
O F: A O=B D: A B .
\end{gathered}
\]

Here,

\[
\begin{gathered}
O F=\varrho, \quad B D=r=8 \\
A B=\sqrt{m^{2}+r^{2}}=\sqrt{225+64}=17 \\
A O=A D+D O=m+(E O-E D)=m+\varrho-x=15+\varrho-x
\end{gathered}
\]

Therefore,

\[
\varrho:(15+\varrho-x)=8: 17 \quad \text { or } \quad \varrho:(15-x)=8: 9
\]

and thus

\[
\varrho=\frac{8(15-x)}{9}
\]

For \( x=0 \), \( \varrho_{1}=\frac{120}{9}=\frac{40}{3} \) is the radius of the sphere \( G_{1} \).

If \( x=2 \varrho_{2} \), then \( \varrho_{2}=\frac{120}{125}=\frac{24}{5}=4.8 \). This is the radius of the sphere \( G_{2} \). Therefore, \( x \) varies in the interval \( 0 \leq x \leq 9.6 \), and here we need to examine the variation of the function

\[
y=\frac{x^{2} \pi}{3}\left(\frac{120-8 x}{3}-x\right)=\frac{\pi}{9}\left(-11 x^{3}+120 x^{2}\right)
\]

We form its first derivative:

\[
y^{\prime}=\frac{\pi}{9}\left(-33 x^{2}+240 x\right)=\frac{\pi}{3} x(-11 x+80)
\]

\( y^{\prime}=0 \) if \( x=0 \). The curve of the function touches the \( X \)-axis at this point.

\( y^{\prime}>0 \) if \( 0 < x < \frac{80}{11} \) and \( y^{\prime}<0 \) if \( x > \frac{80}{11} \). This means that the function, i.e., the volume of the spherical segment, increases as \( x \) increases from 0 to \( \frac{80}{11} \), and then decreases as \( x \) increases from \( \frac{80}{11} \) to \( \frac{48}{5}=9.6 \). If \( x=\frac{80}{11} \), the function has a maximum.[^0]

A table showing some notable values of the function:

| \( x \) | 0 |  | 40  \(\overline{11}\) | 24  \(\overline{5}\) |  | 80  \(\overline{11}\) |  | 48  \(\overline{5}\) |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| \( y \) | 0 | \(\nearrow\) | \(117.55 \pi\) | \(185.1 \pi\) | \(\nearrow\) | \(235.1 \pi\) | \(\searrow\) | \(147.5 \pi\) |

![](https://cdn.mathpix.com/cropped/2024_05_02_8a969ea9e6aa6b332c30g-2.jpg?height=387&width=715&top_left_y=457&top_left_x=694)

Frankl Ottó (Izr. rg. VIII. o. Bp.).

NB. The spherical segment is a hemisphere if \( x=\varrho=\frac{120}{17} \).

This value is close to \( x=\frac{80}{11} \), slightly smaller, so the volume of the spherical segment is slightly larger than that of a hemisphere when it is at its maximum.

Note. If we only examine those spheres that touch the cone's surface, inside the cone, the radius of the larger limiting sphere is perpendicular to \( AB \) at point \( B \). Its size is \( \varrho=\frac{136}{15} \). In this case, the value of \( \varrho \) decreases from \( \frac{136}{15}=9 \frac{1}{15} \) to \( \varrho_{2}=\frac{24}{5}=4.8 \) as \( x \) increases from \( \frac{72}{15}=4.8 \) to \( \frac{48}{5}=9.6 \). The volume of the spherical segment first increases from \( 185.1 \pi \) to the already given maximum, then decreases.

The solution to problem 1291 will be published in the next issue.


[^0]:    \( ^{1} \) The lower vertex of the curve! The axial distance between the lower and upper vertices is bisected by the inflection point, at \( x=\frac{40}{11} \).