Example 2 A glasses workshop in a factory has received a batch of tasks, requiring the processing of 6000 type $A$ parts and 2000 type $B$ parts. This workshop has 214 workers, each of whom can process 3 type $B$ parts in the time it takes to process 5 type $A$ parts. These people are divided into two groups, both working simultaneously, with each group processing one type of part. To complete this batch of tasks in the shortest time possible, how should they be divided?
Solution: Let the number of people assigned to produce $\mathrm{C} A$ type parts be $x$, then the number of people assigned to produce $B$ type parts is $214-x$. Suppose the respective rates are 5 and 3. The time to complete the entire task is
$$
t(x)=\max \left\{\frac{6000}{5 x}, \frac{2000}{3(214-x)}\right\},(x \in N \text {, }
$$
$1 \leqslant x \leqslant 213)$.

When $1 \leqslant x \leqslant 213, x \in N$, the function $\frac{6000}{5 x}$ is decreasing, and the function $\frac{2000}{3(214-x)}$ is increasing.

Thus, $\min t(x)$ is obtained near the solution $x_{0}=137 \frac{4}{7}$ of the equation $\frac{6000}{5 x}=\frac{2000}{3(214-x)}$.

Since $f(137)<f(138)$, the number of people assigned to produce $A$ type parts is 137, and the number of people in the other group is 77.