Example 1 A factory plans to arrange 214 workers to produce 6000 $A$ components and 2000 $B$ components. It is known that the time it takes for each worker to produce 5 $A$ components can produce 3 $B$ components. Now the workers are divided into two groups to produce these two types of components, and they start at the same time. How should the groups be divided to complete the task as quickly as possible?
Solution: Let $x$ be the number of people producing component $A$, then the number of people producing component $B$ is $(214-x)$. Taking the time for one person to produce 5 $A$ components as the unit, the time required to produce 6000 $A$ components and 2000 $B$ components is respectively
$$
y_{1}=\frac{6000}{5 x} \text { and } y_{2}=\frac{2000}{3(214-x)} \text {. }
$$

The total time to complete the task is
$$
T(x)=\max \left\{y_{1}, y_{2}\right\} \text {. }
$$

Note that when $x>0$, the inverse proportion function $y_{1}$ is a decreasing function, and when $0<x<214$, the inverse proportion function $y_{2}$ is an increasing function.
First, consider the equation $y_{1}=y_{2}$, solving it yields $x=137 \frac{4}{7}$.
Then, by the monotonicity of the above $y_{1}$ and $y_{2}$, we get
$$
T_{\min }=\min \{T(137), T(138)\} .
$$

Calculating, we find $T(137)=\frac{6000}{685}<T(138)=\frac{2000}{228}$.
Therefore, arranging 137 people to produce $A$ components and 77 people to produce $B$ components will complete the task the fastest.