11. Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$. As shown in the figure, $F_{1}, F_{2}$ are its left and right foci, $P$ is any point on the ellipse $C$ (not on the endpoints of the major axis), and $PD$ is the angle bisector of $\angle F_{1} P F_{2}$ in $\triangle F_{1} P F_{2}$. Perpendiculars are drawn from $D$ to $P F_{1}$ and $P F_{2}$, with $M$ and $N$ being the feet of these perpendiculars. Try to find: the maximum value of $\frac{S_{\triangle D M N}}{S_{\triangle P F_{1} F_{2}}}$.
11. Solution: Let $\angle F_{1} P F_{2}=\theta$, it is easy to know that $\angle M D N=180^{\circ}-\theta, |D M|=|D N|=|P D| \sin \frac{\theta}{2}$.

From $S_{\triangle P F_{1} F_{2}}=S_{\triangle P F_{1} D}+S_{\triangle P F_{2} D}$, we get $\frac{1}{2}\left|P F_{1}\right|\left|P F_{2}\right| \sin \theta=\frac{1}{2}\left|P F_{1}\right||P D| \sin \frac{\theta}{2}+\frac{1}{2}\left|P F_{2}\right|$ $|P D| \sin \frac{\theta}{2}$. Then $|P D| \sin \frac{\theta}{2}=\frac{\left|P F_{1}\right|\left|P F_{2}\right| \sin \theta}{\left|P F_{1}\right|+\left|P F_{2}\right|}$,
$$
\begin{array}{l}
\frac{S_{\triangle D M N}}{S_{\triangle P F_{1} F_{2}}}=\frac{\frac{1}{2}|D M||D N| \sin \left(180^{\circ}-\theta\right)}{\frac{1}{2}\left|P F_{1}\right|\left|P F_{2}\right| \sin \theta}=\frac{|D M||D N|}{\left|P F_{1}\right|\left|P F_{2}\right|}=\frac{|P D|^{2} \sin ^{2} \frac{\theta}{2}}{\left|P F_{1}\right|\left|P F_{2}\right|}=\frac{\left|P F_{1}\right|\left|P F_{2}\right| \sin ^{2} \theta}{\left(\left|P F_{1}\right|+\left|P F_{2}\right|\right)^{2}} \\
=\frac{1}{4 a^{2}}\left|P F_{1}\right|\left|P F_{2}\right| \sin ^{2} \theta=\frac{1}{4 a^{2}}\left|P F_{1}\right|\left|P F_{2}\right|\left[1-\left(\frac{\left|P F_{1}\right|^{2}+\left|P F_{2}\right|^{2}-4 c^{2}}{2\left|P F_{1}\right|\left|P F_{2}\right|}\right)^{2}\right] \\
=\frac{1}{4 a^{2}}\left|P F_{1}\right|\left|P F_{2}\right|\left(1+\frac{\left|P F_{1}\right|^{2}+\left|P F_{2}\right|^{2}-4 c^{2}}{2\left|P F_{1}\right|\left|P F_{2}\right|}\right)\left(1-\frac{\left|P F_{1}\right|^{2}+\left|P F_{2}\right|^{2}-4 c^{2}}{2\left|P F_{1}\right|\left|P F_{2}\right|}\right) \\
=\frac{1}{8 a^{2}}\left[\left(\left|P F_{1}\right|+\left|P F_{2}\right|\right)^{2}-4 c^{2}\right]\left(2-\frac{\left(\left|P F_{1}\right|+\left|P F_{2}\right|\right)^{2}-4 c^{2}}{2\left|P F_{1}\right|\left|P F_{2}\right|}\right) \\
=\frac{1}{8 a^{2}}\left[4 a^{2}-4 c^{2}\right]\left(2-\frac{4 a^{2}-4 c^{2}}{2\left|P F_{1}\right|\left|P F_{2}\right|}\right)=\frac{b^{2}}{a^{2}}\left(1-\frac{b^{2}}{\left|P F_{1}\right|\left|P F_{2}\right|}\right) \\
\leqslant \frac{b^{2}}{a^{2}}\left(1-\frac{b^{2}}{\left(\frac{\left|P F_{1}\right|+\left|P F_{2}\right|}{2}\right)^{2}}\right)=\frac{b^{2} c^{2}}{a^{4}}, \text { when and only when }\left|P F_{1}\right|=\left|P F_{2}\right|, \text { i.e., } P \text { is the endpoint of the minor axis, the equality }
\end{array}
$$

holds. Therefore, the maximum value sought is $\frac{b^{2} c^{2}}{a^{4}}$.