Example 5. Given a right circular cone with base radius R and height H. Find an inscribed right circular cone of maximum volume (the vertex of the inscribed cone is on the base of the given cone, and the circumference of its base is on the lateral surface of the given cone).





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$$
\begin{array}{l}
\because \mathbf{R} \backslash \triangle A F G \sim R \mathrm{t} \triangle A B D, \\
\therefore-\frac{\mathrm{r}}{\mathbf{R}}=\frac{\mathrm{H}-\mathrm{h}}{\mathrm{H}}, \quad \mathrm{h}=\mathrm{H}\left(1-\frac{\mathrm{r}}{\mathbf{R}}\right),
\end{array}
$$

Therefore, the volume of the inscribed cone is
$$
\begin{array}{l}
\mathrm{V}=\frac{1}{3} \pi \mathrm{H}\left(1-\frac{\mathrm{r}}{\mathrm{H}}\right) \cdot \mathrm{r}^{2} \\
=\frac{\pi H}{3} 4 R^{2}\left(1-\frac{\mathrm{r}}{R}\right) \frac{\mathrm{r}}{2 \mathrm{R}} \cdot \frac{\mathrm{r}}{2 \mathrm{R}}. \\
\text { Since }\left(1-\frac{r}{R}\right)+\frac{r}{2} R^{-}+\stackrel{r}{2 R}=1 \text {, (a constant) } \\
\end{array}
$$

Therefore, when $1--\frac{r}{R}-\frac{r}{2 R}$, i.e., $r=\frac{2}{3} R$,