$18 \cdot 92$ A square $A B C D$ with side length 1, points $E, F$ are on sides $A B$ and $A D$ respectively, and $A E=A F$. If the quadrilateral $C D F E$ has the maximum area, the maximum area is
(A) $\frac{1}{2}$.
(B) $\frac{9}{16}$.
(C) $\frac{19}{32}$.
(D) $\frac{5}{8}$.
(E) $\frac{2}{3}$.
(13th American High School Mathematics Examination, 1962)
[Solution] Let $A E=A F=x$, draw $E G \perp D C$, then
$$
\begin{array}{c}
S_{\triangle E G C}=\frac{1}{2} E G \cdot G C=\frac{1-x}{2}, \\
S_{\text {trapezoid } D F E C}=\frac{1}{2} x[1+(1-x)] \\
=\frac{1}{2} x(2-x), \\
\therefore \quad S_{\text {quadrilateral } C D F E}=S_{\triangle E G C}+S_{\text {trapezoidDFEG }} \\
=\frac{1}{2}\left(1+x-x^{2}\right),
\end{array}
$$

When $x=\frac{1}{2}$, there is a maximum value of $\frac{5}{8}$. Therefore, the answer is $(D)$.