11. (20 points) Given the line $y=x$ intersects the ellipse $C$ : $\frac{x^{2}}{16}+\frac{y^{2}}{11}=1$ at points $A$ and $B$, and a line $l$ passing through the right focus $F$ of the ellipse $C$ with an inclination angle of $\alpha$ intersects the chord $AB$ at point $P$, and intersects the ellipse $C$ at points $M$ and $N$.
(1) Express the area of quadrilateral $M A N B$ in terms of $\alpha$;
(2) Find the equation of the line $l$ when the area of quadrilateral $M A N B$ is maximized.
11. (1) Let the inclination angle of the line $M N$ be $\alpha$, and let $\angle M F O=\theta$. Then
$$
\begin{array}{l}
\alpha+\theta=\pi, \\
|M N|=\frac{2 a b^{2}}{a^{2}-c^{2} \cos ^{2} \alpha}=\frac{2 a b^{2}}{a^{2}-c^{2} \cos ^{2} \theta} .
\end{array}
$$

The angle between $A B$ and $M N$ is $\frac{\pi}{4}+\theta$, so the area of quadrilateral MANB is
$$
\begin{array}{l}
S_{\text {quadrilateral M } 1 N B}=\frac{1}{2}|A B||M N| \sin \left(\frac{\pi}{4}+\theta\right) \\
=\sqrt{2}|O A| a b^{2} \cdot \frac{\sin \theta+\cos \theta}{a^{2}-c^{2} \cos ^{2} \theta} .
\end{array}
$$

Given $a^{2}=16, b^{2}=11, c^{2}=5$,
$$
A\left(\frac{4 \sqrt{33}}{9}, \frac{4 \sqrt{33}}{9}\right) \text {, and }|O A|=\frac{4 \sqrt{66}}{9} \text {, }
$$

Thus, $S_{\text {quadrilateral MANB }}=\frac{352 \sqrt{33}}{9} \cdot \frac{\sin \theta+\cos \theta}{16-5 \cos ^{2} \theta}$
$$
=\frac{352 \sqrt{33}}{9} \cdot \frac{\sin \alpha-\cos \alpha}{16-5 \cos ^{2} \alpha} \text {, }
$$

where $0<\alpha \leqslant \arctan \frac{4 \sqrt{33}}{4 \sqrt{33}+9 \sqrt{5}}$ or
$\arctan \frac{4 \sqrt{33}}{4 \sqrt{33}-9 \sqrt{5}} \leqslant \alpha<\pi$.
(2) Let $f(\alpha)=\frac{\sin \alpha-\cos \alpha}{16-5 \cos ^{2} \alpha}$. The function $f(\alpha)$ can only achieve its maximum value when $\alpha \in\left[\arctan \frac{4 \sqrt{33}}{4 \sqrt{33}-9 \sqrt{5}}, \pi\right)$.
Taking the derivative of $f(\alpha)$, we get
$$
\begin{aligned}
f^{\prime}(\alpha)= & \frac{(\cos \alpha+\sin \alpha)\left(16-5 \cos ^{2} \alpha\right)}{\left(16-5 \cos ^{2} \alpha\right)^{2}}- \\
& \frac{(\sin \alpha-\cos \alpha)(10 \cos \alpha \cdot \sin \alpha)}{\left(16-5 \cos ^{2} \alpha\right)^{2}} .
\end{aligned}
$$

Setting $f^{\prime}(\alpha)=0$, we have
$$
\begin{array}{l}
(1+\tan \alpha)\left(16 \tan ^{2} \alpha+11\right)- \\
(\tan \alpha-1)(10 \tan \alpha)=0 .
\end{array}
$$

Simplifying, we get
$16 \tan ^{3} \alpha+6 \tan ^{2} \alpha+21 \tan \alpha+11=0$.
Thus, $(2 \tan \alpha+1)\left(8 \tan ^{2} \alpha-\tan \alpha+11\right)=0$.
Since $8 \tan ^{2} \alpha-\tan \alpha+11=0$ has no real roots, then $\tan \alpha=-\frac{1}{2}$.

Upon verification, $\tan \alpha=-\frac{1}{2}$ satisfies
$$
\alpha \in\left[\arctan \frac{4 \sqrt{33}}{4 \sqrt{33}-9 \sqrt{5}}, \pi\right] \text {. }
$$

Therefore, the equation of the line $l$ is
$$
y=-\frac{1}{2} x+\frac{\sqrt{5}}{2} .
$$