3. Given that $M$ is a moving point on the arc of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ in the first quadrant, $M N \perp y$ axis, with the foot of the perpendicular being $N$. When the area of $\triangle O M N$ is maximized, the radius $r=$ $\qquad$ of its inscribed circle
3. $\frac{\sqrt{2}}{2}$.

Let $M(4 \cos \theta, 3 \sin \theta)\left(0<\theta<\frac{\pi}{2}\right)$. Then
$$
\begin{array}{l}
S_{\triangle M O N}=\frac{1}{2}|O N||N M| \\
=\frac{1}{2} \times 3 \sin \theta \times 4 \cos \theta=3 \sin 2 \theta .
\end{array}
$$

Therefore, when $\theta=\frac{\pi}{4}$, $\left(S_{\triangle M O N}\right)_{\max }=3$.
At this time, $O N=3 \sin \frac{\pi}{4}=\frac{3 \sqrt{2}}{2}$,
$$
\begin{array}{l}
N M=4 \cos \theta=2 \sqrt{2}, \\
O M=\sqrt{\frac{9}{2}+8}=\frac{5 \sqrt{2}}{2} .
\end{array}
$$

Thus, the inradius of the right triangle $\triangle O M N$ is
$$
r=\frac{1}{2}\left(\frac{3 \sqrt{2}}{2}+2 \sqrt{2}-\frac{5 \sqrt{2}}{2}\right)=\frac{\sqrt{2}}{2} .
$$