2. Find the height of the cone with the largest lateral surface area that can be inscribed in a sphere with radius $R$.
Solution. Let's denote the height of the cone by $x$. The area of the lateral surface of the cone is $y = r s \pi$ (see the diagram). From triangle OAB, we have

$$
\begin{aligned}
& \overline{O A} = x - R \\
& r^{2} = R^{2} - (x - R)^{2} = 2 R x - x^{2}.
\end{aligned}
$$

From triangle $ABC$, we have

$$
s^{2} = x^{2} + r^{2} = x^{2} + 2 R x - x^{2} = 2 R x
$$

from which we get

$$
y = \pi \sqrt{4 R^{2} x^{2} - 2 R x^{3}}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_12f9fb0299f2ef2e3407g-7.jpg?height=440&width=416&top_left_y=1616&top_left_x=1066)

To find the maximum of this function, we take its first derivative:

$$
\begin{aligned}
& y' = \frac{8 R^{2} x - 6 R x^{2}}{2 \sqrt{4 R^{2} x^{2} - 2 R x^{3}}} \pi \\
& y' = 0 \Rightarrow \quad 8 R^{2} x - 6 R x^{2} = 0 \Rightarrow x = 0 \text{ or } x = \frac{8 R^{2}}{6 R} = \frac{4 R}{3}
\end{aligned}
$$

( $x = 0$ is not a solution to the problem);

$$
y'(R) = \pi \frac{R}{\sqrt{2}}, \quad y'\left(\frac{3 R}{2}\right) = -\pi \frac{R}{2}
$$

According to this, the first derivative of the function is positive to the left of the point $x = \frac{4}{3} R$ and negative to the right of that point, which means the function has a maximum at that point.