4. In rectangle $A B C D$, it is known that $A B=3, B C=1$, and a moving point $P$ is on side $C D$. Let $\angle P A B=\alpha, \angle P B A=\beta$, then the maximum value of $\frac{\overrightarrow{P A} \cdot \overrightarrow{P B}}{\cos (\alpha+\beta)}$ is $\qquad$ .
荅事 -3 .
解析 问题转化为求 $|P A| \cdot|P B|$ 的最小值.
由等面积法可得:
$$
\frac{1}{2}|P A| \cdot|P B| \cdot \sin \angle A P B=\frac{1}{2} \times 3 \times 1
$$

所以
$$
|P A| \cdot|P B|=\frac{3}{\sin \angle A P B}
$$

当 $\angle A P B=\frac{\pi}{2}$, 即 $\sin \angle A P B=1$ 时, 所求最大值为 -3 .

Answer -3.
Analysis The problem is transformed into finding the minimum value of $|P A| \cdot|P B|$.
By the equal area method, we get:
$$
\frac{1}{2}|P A| \cdot|P B| \cdot \sin \angle A P B=\frac{1}{2} \times 3 \times 1
$$

Therefore,
$$
|P A| \cdot|P B|=\frac{3}{\sin \angle A P B}
$$

When $\angle A P B=\frac{\pi}{2}$, i.e., $\sin \angle A P B=1$, the maximum value sought is -3.