Example 6 As shown in Figure 7, the side length of square $A B C D$ is $1, P$ is any point on side $B C$ (it can coincide with points $B$ or $C$), and perpendiculars are drawn from points $B, C, D$ to the ray $A P$. The feet of the perpendiculars are $B^{\prime}, C^{\prime}, D^{\prime}$, respectively. Find the maximum and minimum values of $B B^{\prime}+C C^{\prime}+D D^{\prime}$.
Prove that $S_{\triangle A P D}=\frac{1}{2} S_{A B C D}$,
$$
\begin{array}{c}
S_{\triangle A B P}+S_{\triangle A C P}=S_{\triangle A B C}=\frac{1}{2} S_{A B C D} \\
S_{\triangle A B P}+S_{\triangle \triangle A P}+S_{\triangle A P D}=S_{A B C D}=1 \text {, i.e., } \\
\frac{1}{2} A P \cdot B B^{\prime}+\frac{1}{2} A P \cdot C C^{\prime}+\frac{1}{2} A P \cdot D D^{\prime}=1 . \\
A P\left(B B^{\prime}+C C^{\prime}+D D^{\prime}\right)=2 .
\end{array}
$$

Therefore, when $A P$ is the shortest, i.e., when $P$ coincides with $B$, $A P=1$, $B B^{\prime}+C C^{\prime}+D D^{\prime}$ reaches its maximum value of 2; when $A P$ is the longest, i.e., when $P$ coincides with $C$, $A P=\sqrt{2}$, $B B^{\prime}+C C^{\prime}+D D^{\prime}$ reaches its minimum value of $\sqrt{2}$.