11. As shown in Figure 4, the ellipse $\frac{x^{2}}{4}+y^{2}=1$, draw two parallel lines through points $P(1,0)$ and $Q(-1,0)$, intersecting the ellipse at points $A$, $B$, $C$, and $D$. Find the maximum value of the area of quadrilateral $A B C D$.
11. From the conditions and the symmetry of the figure, we know that quadrilateral $ABCD$ is a parallelogram.
Let the slope of line $AB$ be $k$. Then
$$
l_{AB}: y=k(x-1) \text {. }
$$

Substituting into the ellipse equation and simplifying, we get
$$
\left(4 k^{2}+1\right) x^{2}-8 k^{2} x+4\left(k^{2}-1\right)=0 \text {. }
$$

Let the two roots be $x_{A}$ and $x_{B}$, hence
$$
\begin{array}{l}
AB^{2}=\left(1+k^{2}\right)\left(x_{A}-x_{B}\right)^{2} \\
=\left(1+k^{2}\right)\left(\left(x_{A}+x_{B}\right)^{2}-4 x_{A} x_{B}\right) \\
=\left(1+k^{2}\right)\left(\left(\frac{8 k^{2}}{4 k^{2}+1}\right)^{2}-\frac{16\left(k^{2}-1\right)}{4 k^{2}+1}\right) \\
=\frac{16\left(k^{2}+1\right)\left(3 k^{2}+1\right)}{\left(4 k^{2}+1\right)^{2}} . \\
\text { Also, } l_{AB}: y-k x+k=0, \\
l_{CD}: y-k x-k=0,
\end{array}
$$

The distance between the two lines $d=\frac{2|k|}{\sqrt{1+k^{2}}}$.
Thus, $S_{\square ABCD}^{2}=AB^{2} d^{2}=\frac{64 k^{2}\left(3 k^{2}+1\right)}{\left(4 k^{2}+1\right)^{2}}$.
Let $t=4 k^{2}+1$, then $k^{2}=\frac{t-1}{4}$, and $t>1$.
Therefore, $S_{\square ABCD}^{2}=4\left(3-\frac{2}{t}-\frac{1}{t^{2}}\right)<12$
$$
\Rightarrow S_{\square ABCD}<2 \sqrt{3} \text {. }
$$

When $AB \perp x$-axis, it is easy to see that $S_{\square ABCD}=2 \sqrt{3}$.
Thus, $\left(S_{\square ABCD}\right)_{\max }=2 \sqrt{3}$.