5. Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ with its left focus at $F$, the line $y=k x-1$ intersects the ellipse at points $A$ and $B$. When the perimeter of $\triangle F A B$ is maximized, the area of $\triangle F A B$ is $\qquad$
5. $\frac{12 \sqrt{2}}{7}$.

Let the right focus of the ellipse be $C$,
$$
A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right) \text {. }
$$

By the definition of the ellipse,
$$
\left\{\begin{array}{l}
|A F|+|A C|=4, \\
|B F|+|B C|=4 .
\end{array}\right.
$$

Then the perimeter of $\triangle F A B$
$$
\begin{array}{l}
=|A F|+|B F|+|A B| \\
=8+|A B|-(|A C|+|B C|) \\
\leqslant 8,
\end{array}
$$

with equality holding if and only if the line $y=k x-1$ passes through the right focus $C$.
Thus, $k=1$.
Therefore, $S_{\triangle F A B}=\frac{1}{2}|F C|\left|y_{1}-y_{2}\right|$
$$
=\left|y_{1}-y_{2}\right|=\frac{12 \sqrt{2}}{7} \text {. }
$$