11. Let the left focus of the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ be $F$, and let the line $l$ passing through $(1,1)$ intersect the ellipse at points $A$ and $B$. When the perimeter of $\triangle FAB$ is maximized, the area of $\triangle FAB$ is $\qquad$
11.3.

Let $F_{1}$ be the right focus.
Notice,
$$
\begin{array}{l}
F A+F B+A B \\
=2 a-F_{1} A+2 a-F_{1} B+A B \\
=4 a-\left(F_{1} A+F_{1} B-A B\right) \leqslant 4 a,
\end{array}
$$

The equality holds when $l$ passes through point $F_{1}$.
Thus, $l$ is perpendicular to the $x$-axis, at this moment, $S_{\triangle F A B}=3$.