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Does anyone know how to do the third question in junior high school mathematics? Thank you! !
Analysis: (1) annual rent = annual rent per mu of water surface × mu number.

(2) Annual profit = income-cost = (crab seedling income+shrimp seedling income)-(crab seedling cost+shrimp seedling cost)-annual rent on water surface-total feeding cost

(3) Assuming that X mu of water is rented and Y yuan is borrowed from the bank, the annual profit can exceed 35,000 yuan. Firstly, according to the breeding cost ≤ 25,000+Y, Y ≥ 4,900x-25,000 is obtained; From the annual profit of more than 35,000 yuan, 8800 x-60,000 > 1 .08y is obtained, so 8800 x-60,000 >1.08 (4,900 x-25,000) is combined with the meaning of the question to find the value of x: (1.

(2) Income per mu = 4×1400+20×160 = 8800.

The cost per mu = 4× (75+525)+20× (15+85)+500 = 4900.

Profit = 8800-4900 = 3900.

(3) If you rent X mu of water surface and borrow Y yuan from the bank, the annual profit will exceed 35,000 yuan.

Then 500 x+4 (75+525) x+20 (85+15) x ≤ 25000+y,

That is, 4900 x ≤ 25000+y.

y≥4900x-25000

And y≤25000.

( 1400×4+ 160×20)x-[25000+( 1+8%)y]> 35000,

That is, 8800x-60000 >1.08y.

So 8800x-60000 >1.08 (4900x-25000),

The solution is x > 33000/3508 ≈ 9.4,

∵x is an integer,

∴x= 10,

∴y=4900× 10-25000=24000 yuan

That is to rent 10 mu of water surface and borrow 24,000 yuan from the bank.