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
3.
Answer: $\frac{3 \sqrt{11}}{2}$.
Solution: It is easy to know that $A(3,0), F(0,1)$. Let the coordinates of $P$ be $(3 \cos \theta, \sqrt{10} \sin \theta), \theta \in\left(0, \frac{\pi}{2}\right)$, then
$$
\begin{aligned}
S_{\text {OAPF }} & =S_{\triangle O A P}+S_{\triangle O F P}=\frac{1}{2} \cdot 3 \cdot \sqrt{10} \sin \theta+\frac{1}{2} \cdot 1 \cdot 3 \cos \theta \\
& =\frac{3}{2}(\sqrt{10} \cos \theta+\sin \theta)=\frac{3 \sqrt{11}}{2} \sin (\theta+\varphi) .
\end{aligned}
$$

where $\varphi=\arctan \frac{\sqrt{10}}{10}$. When $\theta=\arctan \sqrt{10}$, the maximum area of quadrilateral $O A P F$ is $\frac{3 \sqrt{11}}{2}$.
4.
Answer: 75.
Solution: Consider the steady number $\overline{a b c}$.
If $b=0$, then $a=1, c \in\{0,1\}$, there are 2 steady numbers.
If $b=1$, then $a \in\{1,2\}, c \in\{0,1,2\}$, there are $2 \times 3=6$ steady numbers.
If $2 \leq b \leq 8$, then $a, c \in\{b-1, b, b+1\}$, there are $7 \times 3 \times 3=63$ steady numbers.
If $b=9$, then $a, c \in\{8,9\}$, there are $2 \times 2=4$ steady numbers.
In summary, the number of steady numbers is $2+6+63+4=75$.