11. Given the line $y=x$ intersects the ellipse $C: \frac{x^{2}}{16}+\frac{y^{2}}{11}=1$ at points $A, B$, and a line $l$ passing through the right focus $F$ of the ellipse $C$ with an inclination angle of $\alpha$ intersects the chord $A B$ at point $P$, and intersects the ellipse $C$ at points $M, N$.
(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. Analysis: (1) The inclination angle of the line $M N$ is $\alpha$, and let $\angle M F O=\theta$, where $\angle M F O$ is different from $\alpha$, then $\alpha+\theta=\pi$,
$|M N|=\frac{2 a b^{2}}{a^{2}-c^{2} \cos ^{2} \alpha}=\frac{2 a b^{2}}{a^{2}-c^{2} \cos ^{2} \theta}$. The angle between $A B$ and $M N$ is $\frac{\pi}{4}+\theta$ or $-\frac{3 \pi}{4}+\theta$, then the area of the quadrilateral $M A N B$
$S_{M A N B}=\frac{1}{2}|A B| \cdot|M N|\left|\sin \left(\frac{\pi}{4}+\theta\right)\right|=\sqrt{2}|O A| \cdot a b^{2} \cdot \frac{|\sin \theta+\cos \theta|}{a^{2}-c^{2} \cos ^{2} \theta}$.
Given $a^{2}=16, b^{2}=11, c^{2}=5$, and the coordinates of point $A$ are $\left(\frac{4 \sqrt{33}}{9}, \frac{4 \sqrt{33}}{9}\right)$, and $|O A|=\frac{4 \sqrt{66}}{9}$,
thus, $S_{M A N B}=\frac{352 \sqrt{33}}{9} \cdot \frac{|\sin \theta+\cos \theta|}{16-5 \cos ^{2} \theta}=\frac{352 \sqrt{33}}{9} \cdot \frac{|\sin \alpha-\cos \alpha|}{16-5 \cos ^{2} \alpha}$,
where $0 \leq \alpha<\arctan \frac{4 \sqrt{33}}{4 \sqrt{33}+9 \sqrt{5}}$ or $\arctan \frac{4 \sqrt{33}}{4 \sqrt{33}-9 \sqrt{5}}<\alpha<\pi$.
(2) Let $f(\alpha)=\frac{\sin \alpha-\cos \alpha}{16-5 \cos ^{2} \alpha}$, and $f(\alpha)$ 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: $f^{\prime}(\alpha)=\frac{(\cos \alpha+\sin \alpha)\left(16-5 \cos ^{2} \alpha\right)-(\sin \alpha-\cos \alpha)(10 \cos \alpha \sin \alpha)}{\left(16-5 \cos ^{2} \alpha\right)^{2}}$.
Setting $f^{\prime}(\alpha)=0$, we have $(1+\tan \alpha)\left(16 \tan ^{2} \alpha+11\right)-(\tan \alpha-1)(10 \tan \alpha)=0$.

Simplifying, we get $16 \tan ^{3} \alpha+6 \tan ^{2} \alpha+21 \tan \alpha+11=0$. Therefore, $(2 \tan \alpha+1)\left(8 \tan ^{2} \alpha-\tan \alpha+11\right)=0$.

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 $\alpha \in\left(\arctan \frac{4 \sqrt{33}}{4 \sqrt{33}-9 \sqrt{5}}, \pi\right)$.
Thus, the equation of the line $l$ is: $y=-\frac{1}{2} x+\frac{\sqrt{5}}{2}$.