Example 1 Given $\triangle A B C$ with base $B C=2$, height $A D=1$. Take any point $M$ on $B C$, draw $M N / / A C$ intersecting $A B$ at $N$, and draw $M P / / A B$ intersecting $A C$ at $P$. Try to find where point $M$ is located when the area of $\triangle M N P$ is maximized.
Let $B M=x$. Since $M N / / A C$, we have
$$
M N: A C=x: 2
$$

Since $M P / / B A$, we have

$$
\begin{array}{c}
M P: A B=(2-x): 2 \\
\frac{S_{\triangle M N P}}{S_{\triangle A B C}}=\frac{\frac{1}{2} M N \cdot M P \cdot \sin \angle N M P}{\frac{1}{2} A B \cdot A C \cdot \sin \angle A}=\frac{M N \cdot M P}{A B \cdot A C}=\frac{x(2-x)}{4}
\end{array}
$$

and
$$
S_{\triangle A B C}=\frac{1}{2} B C \cdot A D=1
$$

Thus,
$$
S_{\triangle M N P}=-\frac{1}{4} x^{2}+\frac{1}{2} x=-\frac{1}{4}(x-1)^{2}+\frac{1}{4}
$$

Therefore, when $x=1$, $S_{\triangle M N P}$ reaches its maximum value of $\frac{1}{4}$. Hence, when point $M$ is at the midpoint of $B C$, $S_{\triangle M N P}$ is maximized at $\frac{1}{4}$.