11. (20 points) Let the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ have three moving points $P, Q, R$, and points $P, Q$ are symmetric with respect to the origin. Find the maximum value of $|R P|+|R Q|$.

---

The translation preserves the original text's formatting and line breaks.
11. Let point $P\left(x_{1}, y_{1}\right)$. Then $Q\left(-x_{1},-y_{1}\right)$.

Also, let point $R\left(x_{0}, y_{0}\right)$.
If $x_{0}= \pm x_{1}$, then $R P$ and $R Q$ are parallel to the coordinate axes, respectively, and we have
$$
\begin{array}{l}
|R P|+|R Q|=2\left|x_{1}\right|+2\left|y_{1}\right| . \\
\text { By }\left(\left|x_{1}\right|+\left|y_{1}\right|\right)^{2} \leqslant(16+9)\left(\frac{x_{1}^{2}}{16}+\frac{y_{1}^{2}}{9}\right)=25 \\
\Rightarrow\left|x_{1}\right|+\left|y_{1}\right| \leqslant 5 . \\
\text { Hence }|R P|+|R Q|=2\left|x_{1}\right|+2\left|y_{1}\right| \leqslant 10 .
\end{array}
$$

When the three points are
$$
P\left(\frac{16}{5}, \frac{9}{5}\right), Q\left(-\frac{16}{5},-\frac{9}{5}\right), R\left(\frac{16}{5},-\frac{9}{5}\right)
$$

then, $|R P|+|R Q|=10$.
If $x_{0} \neq \pm x_{1}$, let the line $l_{R P}: y=k x+m$.
Substituting into the ellipse equation and rearranging, we get
$$
\begin{array}{l}
\left(16 k^{2}+9\right) x^{2}+32 k m x+16\left(m^{2}-9\right)=0 \\
\Rightarrow x_{0}+x_{1}=\frac{-32 k m}{16 k^{2}+9}, x_{0} x_{1}=\frac{16\left(m^{2}-9\right)}{16 k^{2}+9} . \\
\text { Hence }|R P|=\sqrt{k^{2}+1}\left|x_{0}-x_{1}\right| \\
=\sqrt{\left(k^{2}+1\right)\left[\left(x_{0}+x_{1}\right)^{2}-4 x_{0} x_{1}\right]} \\
=\frac{24}{16 k^{2}+9} \sqrt{\left(k^{2}+1\right)\left(16 k^{2}-m^{2}+9\right)} .
\end{array}
$$

Now, let the slope of line $R Q$ be $k_{1}$. Then
$$
\begin{array}{l}
k_{1}=\frac{y_{0}+y_{1}}{x_{0}+x_{1}} . \\
\text { By } \frac{x_{0}^{2}}{16}+\frac{y_{0}^{2}}{9}=\frac{x_{1}^{2}}{16}+\frac{y_{1}^{2}}{9} \\
\Rightarrow \frac{x_{0}^{2}-x_{1}^{2}}{16}=-\frac{y_{0}^{2}-y_{1}^{2}}{9} \\
\Rightarrow \frac{y_{0}-y_{1}}{x_{0}-x_{1}} \cdot \frac{y_{0}+y_{1}}{x_{0}+x_{1}}=-\frac{9}{16} \\
\Rightarrow k k_{1}=-\frac{9}{16} \Rightarrow k_{1}=\frac{-9}{16 k} . \\
\text { Hence }|R Q|=\sqrt{1+k_{1}^{2}}\left|x_{0}+x_{1}\right| \\
=\sqrt{\left[1+\left(\frac{-9}{16 k}\right)^{2}\right]}\left|\frac{-32 k m}{16 k^{2}+9}\right| \\
=\frac{2}{16 k^{2}+9} \sqrt{16^{2} k^{2}+9^{2}}|m| .
\end{array}
$$
(1) + (2) and combining with the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
|R P|+|R Q| \\
=\frac{24}{16 k^{2}+9} \sqrt{\left(k^{2}+1\right)\left(16 k^{2}-m^{2}+9\right)}+ \\
\frac{2}{16 k^{2}+9} \sqrt{16^{2} k^{2}+9^{2}}|m| \\
=\frac{24}{16 k^{2}+9}\left[\sqrt{\left(k^{2}+1\right)\left(16 k^{2}-m^{2}+9\right)}+\sqrt{\left(\frac{16}{9} k^{2}+\frac{9}{16} m^{2}\right.}\right] \\
\leqslant \frac{24}{16 k^{2}+9} \sqrt{\left(k^{2}+1+\frac{16}{9} k^{2}+\frac{9}{16}\right)\left(16 k^{2}-m^{2}+9+m^{2}\right)} \\
=10 .
\end{array}
$$

Therefore, the maximum value of $|R P|+|R Q|$ is 10.