Then if the second person works regularly for (t- 12) days, his profit is 50*(t- 12).
It is concluded that the profit created by m- 1 person in t days is 50 * (t-12-6-4-12/(m-2)).
Find the total profit of m- 1 person in t days and get a series. You can find it yourself. I'm lazy, so I won't. Suppose it is found to be S(m- 1).
It is also assumed that the recruitment will be stopped when there are just m workers in the factory, that is, at this time, S(m- 1)=600*(m- 1), that is, there is no profit accumulation and surplus when m workers are recruited, and 20,000 workers in this factory will be completed in less than 12 days. 12. And if we recruit one more person at this time, the task will be delayed. In this case, the task should be completed within y days after the recruitment is stopped (y can also be a decimal), so these inequalities hold:
y & gtx
(m+ 1)*50*y=20000
In fact, this becomes a problem of finding the optimal solutions of three unknowns and two constrained equations.
Equation 1: 20000=m*50*x
Equation 2: 20000=(m+ 1)*50*y
The constraint conditions are: x
The optimal solution for solving x is x=? And find m=? (Note: m must be a positive integer. ) No, it's hard to calculate.
Then bring m in:
S(m- 1)=600*(m- 1)
Get t=？ , the total number of days is t+x days.
Brother, this question is really unusual! ! ! ! ! ! I don't know if I have solved it correctly, but I won't work out the answer anyway. Why don't you do the math and see if it's right?