11、Given the ellipse $C: \frac{x^{2}}{3}+y^{2}=1$ with its upper vertex $A$, a line $l$ that does not pass through $A$ intersects the ellipse $C$ at points $P$ and $Q$, and $A P \perp A Q$. Find the maximum value of the area of $\triangle A P Q$.

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11. Analysis: When the slope of line $P Q$ does not exist, we know that $\angle P A Q$ is an acute angle, which does not meet the problem's requirements. Therefore, we only need to consider the case where the slope exists.

Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, and the equation of line $P Q$ be $y=k x+b$. By combining the ellipse equation $\frac{x^{2}}{3}+y^{2}=1$, and eliminating $y$, we get $\left(k^{2}+\frac{1}{3}\right) x^{2}+2 k b x+b^{2}-1=0$. Thus, $x_{1}+x_{2}=\frac{-2 k b}{k^{2}+\frac{1}{3}}, x_{1} x_{2}=\frac{b^{2}-1}{k^{2}+\frac{1}{3}}$. According to the problem, $A P \perp A Q$, so $A P \cdot A Q=\stackrel{1}{0}$, which means $x_{1} x_{2}+\left(y_{1}-1\right)\left(y_{2}-1\right)=0$. Therefore,
$$
\begin{array}{l}
0=x_{1} x_{2}+\left(y_{1}-1\right)\left(y_{2}-1\right)=x_{1} x_{2}+\left(k x_{1}+b-1\right)\left(k x_{2}+b-1\right)=\left(k^{2}+1\right) x_{1} x_{2}+k(b-1)\left(x_{1}+x_{2}\right)+(b-1)^{2} \\
=\frac{\left(k^{2}+1\right)\left(b^{2}-1\right)}{k^{2}+\frac{1}{3}}-\frac{2 k b(b-1)}{k^{2}+\frac{1}{3}}+(b-1)^{2}=\frac{\left(k^{2}+1\right)\left(b^{2}-1\right)-2 k b(b-1)+(b-1)^{2}\left(k^{2}+\frac{1}{3}\right)}{\left(k^{2}+\frac{1}{3}\right)},
\end{array}
$$

Thus, $\left(k^{2}+1\right)\left(b^{2}-1\right)-2 k b(b-1)+(b-1)^{2}\left(k^{2}+\frac{1}{3}\right)=0$, simplifying to $b=-\frac{1}{2}$. Therefore, the equation of line $P Q$ is $y=k x-\frac{1}{2}$, which always passes through the fixed point $C\left(0,-\frac{1}{2}\right)$. We have $x_{1}+x_{2}=\frac{3 k}{3 k^{2}+1}, x_{1} x_{2}=\frac{-9}{4\left(3 k^{2}+1\right)}$, thus
$$
\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}=\sqrt{\left(\frac{3 k}{3 k^{2}+1}\right)^{2}+\frac{9}{\left(3 k^{2}+1\right)}}=\frac{6 \sqrt{k^{2}+\frac{1}{4}}}{3 k^{2}+1} \text {. }
$$

Let $t=\sqrt{k^{2}+\frac{1}{4}}$, then $t \geq \frac{1}{2} \cdot S_{\triangle A P Q}=\frac{1}{2}|A C| \cdot\left|x_{1}-x_{2}\right|=\frac{9 t}{3 t^{2}+\frac{1}{4}}=\frac{9}{3 t+\frac{1}{4 t}}\left(t \geq \frac{1}{2}\right)$, we know that $t=\frac{1}{2}$, i.e., $k=0$, when $S_{\triangle A P Q}$ reaches its maximum value $\frac{9}{4}$. Therefore, the maximum area of $\triangle A P Q$ is $\frac{9}{4}$.