7. In $\square A B C D$, $A B=4, B C=2, \angle D A B$ $=60^{\circ}$, a point $P$ on side $B C$ and a point $Q$ on the extension of $D C$ satisfy $S_{\triangle B P Q}=\frac{\sqrt{3}}{32}$. Then the minimum value of $\overrightarrow{A P} \cdot \overrightarrow{B Q}$ is $\qquad$
7. $\frac{97}{16}$.

As shown in Figure 2.
Let $\overrightarrow{A B}=\boldsymbol{a}, \overrightarrow{A D}=\boldsymbol{b}, \overrightarrow{C Q}=u \boldsymbol{a}, \overrightarrow{B P}=v \boldsymbol{b}$, where the angle between vectors $\boldsymbol{a}, \boldsymbol{b}$ is $60^{\circ}$, and $u, v > 0$.
It is easy to know that the distance from point $Q$ to line $B C$ is $\frac{\sqrt{3}}{2} C Q$.
$$
\begin{array}{l}
\text { Then } \frac{\sqrt{3}}{32}=S_{\triangle B P Q}=\frac{\sqrt{3}}{4} B P \cdot C Q \\
=\frac{\sqrt{3}}{4} v A D \cdot u A B=2 \sqrt{3} u v \\
\Rightarrow u v=\frac{1}{64} . \\
\text { Therefore, } \overrightarrow{A P} \cdot \overrightarrow{B Q}=(\boldsymbol{a}+v \boldsymbol{b})(\boldsymbol{b}+u \boldsymbol{a}) \\
=u \boldsymbol{a}^{2}+v \boldsymbol{b}^{2}+(1+u v) \boldsymbol{a} \cdot \boldsymbol{b} \\
=16 u+4 v+(1+u v) \times 4 \times 2 \cos 60^{\circ} \\
=4(4 u+v)+\frac{65}{16} \\
\geqslant 8 \sqrt{4 u v}+\frac{65}{16}=\frac{97}{16} .
\end{array}
$$

When and only when $(u, v)=\left(\frac{1}{16}, \frac{1}{4}\right)$, the minimum value of $\overrightarrow{A P} \cdot \overrightarrow{B Q}$ is $\frac{97}{16}$.