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Mathematics of environmental protection
1

1. Single purchase of Type A can treat sewage every month: 240 tons/month * 10 sets = 2,400 tons.

Cost: 12* 10 set+1* 10 set = 65438+ 0.3 million over budget, and we can't just buy model A.

2. Buying only Type B can treat sewage every month: 200 tons/month * 10 sets = 2,000 tons, so you can't just buy Type B..

3. buy x sets of type a, 10-x sets of type b, with a total price of y.

The solution of 240x+200( 10-x)≥2040 is x≥ 1.

Y =12x+10 (10-x) y = 2x+100, and the total price is the least. X is 1, and we bought 1 model a and 9 model b. The total price is 6,543,800 yuan+0,200 yuan.

2

Let the cost in the n(n≤ 15) year be equal to the sewage treatment cost,

N-year sewage treatment equipment cost = (102/15) * n+1*10 * n =1.68 billion yuan.

N-year sewage treatment cost of sewage plant = 10 yuan/ton *2040 ton/month * 12n month = 244,800 yuan = 244,800 yuan.

16.8n=24.48n, the solution is n=0,

That is, it is more economical to treat each ton of water from the beginning than a sewage treatment plant.