14. Given a line segment $C D$ of length 6 with midpoint $M$. Two triangles $\triangle A C D$ and $\triangle B C D$ are constructed on the same side with $C D$ as one side, and both have a perimeter of 16, satisfying $\angle A M B = 90^{\circ}$. Find the minimum area of $\triangle A M B$.
14. As shown in Figure 7, points $A$ and $B$ move on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$ (with $C$ and $D$ being the left and right foci, respectively).
Let $l_{M A}: y=k x$. Then,
$$
\left\{\begin{array}{l}
y = k x, \\
\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1
\end{array} \Rightarrow \left\{\begin{array}{l}
x_{A}^{2} = \frac{25 \times 16}{16 + 25 k^{2}}, \\
y_{A}^{2} = k^{2} x_{A}^{2}.
\end{array}\right.\right.
$$

Thus, $|M A|^{2} = x_{A}^{2} + y_{A}^{2}$
$$
= \left(1 + k^{2}\right) \frac{25 \times 16}{16 + 25 k^{2}}.
$$

By substituting $-\frac{1}{k}$ for $x$, we get
$$
\begin{array}{l}
|M B|^{2} = \left[1 + \left(-\frac{1}{k}\right)^{2}\right] \frac{25 \times 16}{16 + 25\left(-\frac{1}{k}\right)^{2}} \\
= \left(1 + k^{2}\right) \frac{25 \times 16}{16 k^{2} + 25}.
\end{array}
$$

Therefore, $S_{\triangle M M B}^{2} = \frac{1}{4} |M A|^{2} \cdot |M B|^{2}$
$$
\begin{array}{l}
= \frac{1}{4} \times 25 \times 16 \times \frac{25 \times 16 \left(1 + k^{2}\right)^{2}}{\left(16 k^{2} + 25\right) \left(16 + 25 k^{2}\right)} \\
= \frac{1}{4} \times 25 \times 16 \left[1 - \frac{81}{25 \times 16 k^{2} + \frac{25 \times 16}{k^{2}} + \left(25^{2} + 16^{2}\right)}\right] \\
\geqslant 100 \left[1 - \frac{81}{2 \sqrt{25 \times 16 k^{2} \cdot \frac{25 \times 16}{k^{2}}} + \left(25^{2} + 16^{2}\right)}\right] \\
= 100 \times \frac{40^{2}}{41^{2}}.
\end{array}
$$

Thus, $\left(S_{\triangle A M B}\right)_{\min} = \frac{400}{41}$.
The minimum value is achieved when $k = \pm 1$.