4. In rectangle $A B C D$, $A B=2, A D=1$, point $P$ on segment $D C$ and point $Q$ on the extension of $C B$ satisfy the condition $|\overrightarrow{D P}|=|\overrightarrow{B Q}|$, then the minimum value of $\overrightarrow{P A} \cdot \overrightarrow{P Q}$ is $\qquad$ .
The answer is $\frac{3}{4}$. Solution: Let's set $\mathrm{A}(\mathrm{o}, \mathrm{o}), \mathrm{B}(2, \mathrm{o}), \mathrm{D}(\mathrm{o}, \mathrm{l})$. Suppose the coordinates of $\mathrm{P}$ are $(t, \mathrm{l})$ (where $0 \leq t \leq 2$), then from $|\overrightarrow{D P}|=|\overrightarrow{B Q}|$ we get the coordinates of $\mathrm{Q}$ as $(2,-t)$, hence $\overrightarrow{P A}=(-t,-1), \overrightarrow{P Q}=(2-t,-t-1)$, therefore,
$$
\begin{array}{l}
\overrightarrow{P A} \cdot \overrightarrow{P Q}=(-t) \cdot(2-t)+(-1) \cdot(-t-1)=t^{2}-t+1=\left(t-\frac{1}{2}\right)^{2}+\frac{3}{4} \geq \frac{3}{4} . \\
\text { When } t=\frac{1}{2} \text {, }(\overrightarrow{P A} \cdot \overrightarrow{P Q})_{\min }=\frac{3}{4} .
\end{array}
$$