15. As shown in Figure $2, A$ is a fixed point on the circle $\odot O$ with radius 1, and $l$ is the tangent line of $\odot O$ passing through point $A$. Let $P$ be a point on $\odot O$ different from $A$, and $P Q \perp$ $l$, with the foot of the perpendicular being $Q$. When point $P$ moves on $\odot O$, the maximum value of the area of $\triangle P A Q$ is $\qquad$
15. $\frac{3 \sqrt{3}}{8}$.

As shown in Figure 9, let the symmetric point of point $P$ with respect to line $AO$ be $P'$. Then,
$$
S_{\triangle PAQ}=\frac{1}{2} S_{\triangle PAP} .
$$

By plane geometry knowledge, when and only when $\triangle PAP'$ is an inscribed equilateral triangle in the circle, the area of $\triangle PAP'$ is maximized, with the maximum value being $\frac{3 \sqrt{3}}{4}$. Therefore, the maximum area of $\triangle PAQ$ is $\frac{3 \sqrt{3}}{8}$.