3. Given the ellipse $\frac{x^{2}}{5}+\frac{y^{2}}{4}=1$ with its right focus at $F$, the upper vertex at $A$, and $P$ as a moving point on the ellipse in the first quadrant. Then the maximum value of the area of $\triangle A P F$ is $\qquad$ .
3. $\sqrt{6}-1$.

From the problem, we know $A(0,2), F(1,0)$.
Thus, $|A F|=\sqrt{5}$.
Let $P(\sqrt{5} \cos \alpha, 2 \sin \alpha)\left(\alpha \in\left(0, \frac{\pi}{2}\right)\right)$.
Then the distance from point $P$ to the line $l_{A F}: x+\frac{y}{2}=1$ is
$$
\begin{array}{l}
h=\frac{|2 \sqrt{5} \cos \alpha+2 \sin \alpha-2|}{\sqrt{5}} . \\
\text { Therefore, } S_{\triangle A P F}=\frac{1}{2}|A F| h \\
=|\sqrt{5} \cos \alpha+\sin \alpha-1| \\
=|\sqrt{6} \sin (\alpha+\varphi)-1| \leqslant \sqrt{6}-1 .
\end{array}
$$