Let the initial growth rate of unit grassland be a(N cubic meters/week/hectare)

Let the grazing speed of unit cattle be b(N cubic meters/week/head)

Assume that the original grassland per hectare is 1(N cubic meters).

The growth acceleration (deceleration) of the first grassland is 5/6a.

12*4b= 10/3+ 10/3a*4- 1/2*5/6a*4^2

The growth acceleration (deceleration) of the second grassland is 10/9a.

2 1*9b= 10+ 10a*9- 1/2* 10/9a*9^2

The solution is a= 1/6 b=5/54.

The relevant parameters of the third grassland are as follows

x* 18*5/54=24+24* 1/6* 18- 1/2*2/9* 18^2

Get x=36.

So, you can feed 36 cows, hahaha, it's worked out. Directly participate in the physical acceleration formula, using the mathematical limit thought. It's really hard. Calculation and setting are troublesome. I have been calculating for a long time. If you don't choose me as the answer, you will regret it.

The key to solve the problem is that the growth rate of unit grassland is not affected by the number of cattle, that is, no matter how many cattle come to eat grass, the growth rate of grass will not change. Some units I set up are for thinking, and do not represent the actual figures.

If there is grass on the eaten part, it is

Suppose a cow eats X once a week and grows Y once a hectare. rule

12x×4 = 10/3( 1+4y)

2 1x×9 = 10( 1+9y)

zx× 18 = 24( 1+ 18y)

Solution: z=36

Answer: 36 cows ate the third block 18 weeks.

Why is there a difference? The answer is still the same, because in the previous solution, the speed of cattle eating slowed down. It can be seen that no matter how long the grass grows, the answer has no effect, which is also a feature of contrast. Therefore, when solving similar problems in the future, you can idealize the questions first and then answer them.

I answered on the 27th, and today on the 30th, please change it again. )