18.93 As shown in the figure, a square with a side length of 4 is cut at one corner to form a pentagon $A B C D E$, where $A F=2, F B=1$. Find a point $P$ on $A B$ such that the rectangle $P N D M$ has the maximum area, then the position of point $P$ is
(A) at point $A$.
(B) at the midpoint of $A B$.
(C) at the point where $\frac{A P}{P B}=2$.
(D) at point $B$.
(Chinese Jiangsu Province Xuzhou Junior High School Mathematics Competition, 1988)
[Solution] Extend $M P, N P$ to intersect $F C, E F$ at $G, K$, respectively, and let $P K=x$, then
$$
\begin{array}{c}
B G=1-x . \\
\because \triangle B G P \backsim \triangle B F A, \therefore \frac{P G}{B G}=\frac{A F}{B F}=2,
\end{array}
$$

which means
$$
P G=2(1-x) .
$$

and $\quad M P=4-2(1-x)$.
$$
\text { Also, } \begin{aligned}
S_{P \mathrm{NDM}} & =M P \cdot P N=(4-x)[4-2(1-x)] \\
& =-2 x^{2}+6 x+8,
\end{aligned}
$$

When $0 \leqslant x \leqslant 1$, the maximum value of $S_{P \text { NDM }}$ is 12.
That is, when $P$ coincides with $B$, $S_{P N D M}$ is maximized.
Therefore, the answer is $(D)$.