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Mathematical modeling. Help!
If we only consider these two months, the problem will be simple! (Because there is no need to consider the storage capacity and profit maximization after the third month)

The specific method is as follows:

Since A and B can generate up to 60,000 and 35,000 electric power per month, A and B can use up to 150M and 175M of water per month.

(60000÷400= 150 35000÷200= 175)

According to the topic, A can store 1900+200=2 100 and 2 100+ 130=2230 at most within two months.

Moreover, because A stores at least 1800M of water every month, A can use 150M of water every two months.

It can be seen from the figure that the water used by A can be used as the water source of B.

So B can deposit 850+40+ 150= 1040 and140+15+150 =1205 at most in two months.

Because B stores at least 800M water every month, B can use 175M water every two months.

Therefore, it can be concluded that the maximum monthly income of the power plant = 50,000× 200+(60,000+35,000-50,000 )×140 =16.3 million yuan.

The modeling process is as follows:

Let A use water this month, X use water next month, Y use water this month, B use water next month, N use water next month, and the income of the power plant for two months is R, R.

Question (Power station A can convert 1000 M3 of water in reservoir A into 400 kWh of electric energy, and power station B can convert 1000 m3 of water in reservoir B into electric energy.

M3 of water is converted into 200 kWh of electricity. Power station A, the maximum monthly power generation is 60,000 kWh and 35,000 kWh respectively.

Degree) can get x, y.

According to my analysis above (comparing the minimum storage capacity), we can get X >: = 100(2 100 exceeds the maximum value 100).

Y & gt=80 (assuming that A used the most water of 150M last month, then Y can take the minimum value of1950+130-2000 = 80m) m, and n > =0.

If 400x+200m, then r1= 200x (400x+200m).

R2 = 200× 50000+140× (400x+200m-50000) If 400x+200m >; 50000

Similarly, r1= 200× (400y+200n) if 400y+200n.

R2 = 200× 50000+140× (400y+200n-50000) If 400y+200n >; 50000

According to my analysis above, XY can be 150 mn, and both xy can be 175.

Therefore, according to the principle of maximizing the income in two months, the manufacturer will choose X, Y = 150m and N = 175.

R2= 16300000

r2= 16300000

Dinner, don't talk like that. Who will do all the questions as soon as they come up? Isn't it normal not to ask others? This is the spirit of reading! !