3. Given a square $A B C D$ with side length $1, E$ and $F$ are points on sides $B C$ and $C D$ respectively, and satisfy $B E=C F$. Then the minimum value of the area of $\triangle A E F$ is ( ).
(A) $\frac{\sqrt{2}}{8}$
(B) $\frac{\sqrt{3}}{8}$
(C) $\frac{1}{4}$
(D) $\frac{3}{8}$
3. D.

Let $B E=C F=x$. Then $C E=D F=1-x$, and
$$
\begin{array}{l}
S_{\triangle A E F}=1-\frac{1}{2} x-\frac{1}{2}(1-x)-\frac{1}{2} x(1-x) \\
=\frac{1}{2}\left[\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}\right] \geqslant \frac{3}{8} .
\end{array}
$$

When $x=\frac{1}{2}$, the equality holds.
Therefore, the minimum value of the area of $\triangle A E F$ is $\frac{3}{8}$.