Example 3: After long-term observation, it is found that in a busy traffic period, the traffic flow $y$ (thousand vehicles/hour) of a certain road section and the average speed $v$ (km/h) of vehicles have the following functional relationship:
$$
y=\frac{920 v}{v^{2}+3 v+1600}(v>0) .
$$
(1) In this period, at what average speed $v$ is the traffic flow maximum? What is the maximum traffic flow? (Accurate to 0.1 thousand vehicles/hour)
(2) If the traffic flow needs to exceed 10 thousand vehicles/hour in this period, what should be the range of the average speed of vehicles?
Solution (1) According to the problem, $y=\frac{920}{3+\left(v+\frac{1600}{v}\right)} \leqslant \frac{920}{3+2 \sqrt{1600}}=\frac{920}{83}$, equality holds if and only if $v=\frac{1600}{v}$, i.e., $v=40$.
Therefore, $y_{\text {max }}=\frac{920}{83} \approx 11.1$.
(2) From the condition, we get $\frac{920 v}{v^{2}+3 v+1600}>10$, rearranging gives

i.e.,
$$
\begin{array}{l}
v^{2}-89 v+1600<0, \\
(v-25)(v-64)<0 .
\end{array}
$$

Solving gives $25<v<64$.
Answer: (1) When the average speed of the car $v=40$ (km/h), the traffic flow is maximized, with the maximum traffic flow being approximately 11.1 thousand vehicles/hour; (2) If the traffic flow is required to exceed 10 thousand vehicles/hour during this period, the average speed of the car should be greater than 25 km/h and less than 64 km/h.

Solution Guidance: In solving the maximum value of the traffic flow, the basic inequality for the maximum value was used, and the equality can be exactly achieved. If the equality cannot be achieved here, the monotonicity of the function $f(v)=v+\frac{a}{v}$ about $v$ (content of the next chapter) can be utilized.