3. In the Cartesian coordinate system $x O y$, the equation of the ellipse $C$ is $\frac{x^{2}}{9}+\frac{y^{2}}{10}=1, F$ is the upper focus of $C$, $A$ is the right vertex of $C$, and $P$ is a moving point on $C$ located in the first quadrant. Then the maximum value of the area of quadrilateral $O A P F$ is $\qquad$.
Let $P(3 \cos \theta, \sqrt{10} \sin \theta), \theta \in\left(0, \frac{\pi}{2}\right), A(3,0), F(0,1)$,
then $S_{O A P F}=S_{\triangle P O A}+S_{\triangle P O F}$
$=\frac{1}{2}|O A| \cdot y_{P}+\frac{1}{2}|O F| \cdot x_{P}=\frac{3}{2}(\sqrt{10} \sin \theta+\cos \theta)$
$=\frac{3}{2} \sqrt{11} \sin (\theta+\varphi) \leqslant \frac{3}{2} \sqrt{11}$. Equality holds when
$\sin \theta=\frac{\sqrt{110}}{11}, \cos \theta=\frac{\sqrt{11}}{11}$, i.e., $P\left(\frac{3 \sqrt{11}}{11}, \frac{10 \sqrt{11}}{11}\right)$.
Therefore, the maximum area of quadrilateral $O A P F$ is $\frac{3}{2} \sqrt{11}$.