14. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with an eccentricity of $\frac{1}{2}$, $F_{1}$ and $F_{2}$ are the left and right foci, respectively. A line passing through $F_{2}$ intersects the ellipse at points $A$ and $B$. If the maximum area of $\triangle F_{1} A B$ is 6, find the equation of the ellipse.
14. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$,

line $l_{A B}: x=m y+c$.
Substituting it into the ellipse equation yields ...
$$
\left(a^{2}+b^{2} m^{2}\right) y^{2}+2 b^{2} c m y-b^{4}=0 \text {. }
$$

Thus, $S_{\triangle F_{1} A B}=\frac{1}{2}\left|F_{1} F_{2}\right|\left|y_{1}-y_{2}\right|$
$$
=c \cdot \frac{\sqrt{\Delta}}{a^{2}+b^{2} m}=\frac{2 a b^{2} c \sqrt{1+m^{2}}}{c^{2}+b^{2}\left(1+m^{2}\right)} \text {. }
$$

Given $e=\frac{c}{a}=\frac{1}{2}$, we have
$$
a=2 c, b=\sqrt{a^{2}-c^{2}}=\sqrt{3} c \text {. }
$$

Then, $S_{\triangle F_{1} A B}=\frac{12 c^{2} \sqrt{1+m^{2}}}{1+3\left(1+m^{2}\right)}$
$$
=4 c^{2}\left(\sqrt{1+m^{2}}+\frac{1}{3 \sqrt{1+m^{2}}}\right)^{-1} \text {. }
$$

And $\sqrt{1+m^{2}}+\frac{1}{3 \sqrt{1+m^{2}}}$
$$
\begin{array}{l}
=\frac{2}{3} \sqrt{1+m^{2}}+\frac{1}{3}\left(\sqrt{1+m^{2}}+\frac{1}{\sqrt{1+m^{2}}}\right) \\
\geqslant \frac{2}{3}+\frac{1}{3} \times 2=\frac{4}{3} .
\end{array}
$$

Therefore, $S_{\triangle F_{1} A B} \leqslant 4 c^{2} \times \frac{3}{4}=3 c^{2}$.
When and only when $m=0$, $S_{\triangle F 1 A B}=3 c^{2}$.
Thus, $3 c^{2}=6, c^{2}=2, a^{2}=8, b^{2}=6$.
Hence, the required equation is $\frac{x^{2}}{8}+\frac{y^{2}}{6}=1$.