11. (20 points) Given the ellipse $\Gamma$:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0),
$$
$A$ and $B$ are the left vertex and the upper vertex of the ellipse $\Gamma$, respectively, and $P$ is a point on the ellipse $\Gamma$ in the fourth quadrant. $PA$ intersects the $y$-axis at point $C$, and $PB$ intersects the $x$-axis at point $D$. Find the maximum value of the area of $\triangle PCD$.
11. From the problem, we get $A(-a, 0), B(0, b)$. Let $P(m, n)(m>0, n<0)$. Then
$$
\begin{array}{l}
l_{P A}: y=\frac{n}{m+a}(x+a), \\
l_{P B}: y=\frac{n-b}{m} x+b .
\end{array}
$$

Thus, $C\left(0, \frac{n a}{m+a}\right), D\left(\frac{m b}{b-n}, 0\right)$.
Hence, $|A D|=\frac{m b}{b-n}+a=\frac{m b+a b-n a}{b-n}$,
$S_{\triangle P C D}=\frac{1}{2}|A D|\left|y_{C}-y_{P}\right|$
$=\frac{1}{2} \cdot \frac{m b+a b-n a}{b-n} \cdot \frac{-m n}{m+a}$
$=\frac{1}{2} \cdot \frac{m^{2} n b+m n a b-m n^{2} a}{(n-b)(m+a)}$,
$S_{\triangle P C D}=\frac{1}{2} \cdot \frac{n b\left(a^{2}\left(1-\frac{n^{2}}{b^{2}}\right)\right)+m n a(b-n)}{(n-b)(m+a)}$
$=\frac{1}{2} \cdot \frac{n b\left(\frac{a^{2}}{b^{2}}\left(b^{2}-n^{2}\right)\right)+m n a(b-n)}{(n-b)(m+a)}$
$=-\frac{1}{2} \cdot \frac{n b\left(\frac{a^{2}}{b^{2}}(b+n)\right)+m n a}{m+a}$
$=-\frac{1}{2} \cdot \frac{n a^{2}+\frac{n^{2} a^{2}}{b}+m n a}{m+a}$
$=-\frac{1}{2} \cdot \frac{n a^{2}+\frac{a^{2} b^{2}-b^{2} m^{2}}{b}+m n a}{m+a}$
$=-\frac{1}{2} \cdot \frac{n a(a+m)+b(a+m)(a-m)}{m+a}$
$=\frac{b m-a n-a b}{2}$.
Let $m=a \cos \theta, n=b \sin \theta$. Then
$z=b m-a n-a b$
$=a b \cos \theta-a b \sin \theta$
$=a b(\cos \theta - \sin \theta)-a b$
$=-\sqrt{2} a b \sin \left(\theta-\frac{\pi}{4}\right)-a b$
$\leqslant(\sqrt{2}-1) a b$,
with equality holding if and only if $\theta-\frac{\pi}{4}=-\frac{3 \pi}{2}$.
Therefore, the maximum area of $\triangle P C D$ is $\frac{\sqrt{2}-1}{2} a b$.