11. (20 points) In the Cartesian coordinate system $x O y$, $A$ is a point on the ellipse $\frac{x^{2}}{4}+y^{2}=1$, $M$ is a moving point on the line segment $O A$, and a line through $M$ intersects the ellipse at points $P$ and $Q$. If $\overrightarrow{P M}=2 \overrightarrow{M Q}$, find the maximum value of the area of quadrilateral $O P A Q$.
11. As shown in Figure 5.

Let \( P(2 \cos \alpha, \sin \alpha), Q(2 \cos \beta, \sin \beta) \).
Since \( \overrightarrow{P M}=2 \overrightarrow{M Q} \), we have
\[
x_{M}=\frac{2 \cos \alpha+4 \cos \beta}{3}, y_{M}=\frac{\sin \alpha+2 \sin \beta}{3} \text{. }
\]

Let \( \overrightarrow{O A}=u \overrightarrow{O M} (u>1) \). Then
\[
\begin{array}{l}
x_{A}=\frac{u}{3}(2 \cos \alpha+4 \cos \beta), \\
y_{A}=\frac{u}{3}(\sin \alpha+2 \sin \beta) .
\end{array}
\]

Substituting into the ellipse equation, we get
\[
\begin{array}{l}
\frac{\left(\frac{u}{3}\right)^{2}(2 \cos \alpha+4 \cos \beta)^{2}}{4}+\left(\frac{u}{3}\right)^{2}(\sin \alpha+2 \sin \beta)^{2}=1 \\
\Rightarrow 5+4 \cos (\alpha-\beta)=\frac{9}{u^{2}} \\
\Rightarrow \cos (\alpha-\beta)=\frac{9}{4 u^{2}}-\frac{5}{4} .
\end{array}
\]

Thus, \( S=u S_{\triangle P O Q} \)
\[
\begin{array}{l}
=u \cdot \frac{1}{2}|2 \cos \alpha \cdot \sin \beta-2 \cos \beta \cdot \sin \alpha| \\
=u|\sin (\alpha-\beta)| \\
=u \sqrt{1-\left(\frac{9}{4 u^{2}}-\frac{5}{4}\right)^{2}} \\
=\sqrt{-\left(\frac{9}{16} u^{2}+\frac{81}{16 u^{2}}\right)+\frac{45}{8}} \leqslant \frac{3}{2} .
\end{array}
\]

The equality holds when \( u=\sqrt{3} \).
Therefore, the maximum value of the area \( S \) is \( \frac{3}{2} \).