## sritama 3 years ago Suppose you have a grassy field, and cows eat grass at a constant rate. Keep in mind, the grass keeps growing continuously. 48 cows can clear all the grass off the field in 90 days. 120 cows can clear all the grass off the field in 30 days. How many cows would be needed to clear all of the grass in 16 days? Round up to the nearest whole cow.

1. experimentX

48 cows can clear all the grass off the field in 90 days. 120 cows can clear all the grass off the field in 30 days. ??

2. sritama

yeah...

3. gingerkid101

no contradiction. the grass keeps growing. \[initial + (grassgrowthrate * days) = (numcows * eatingrate) \] for simplification x + y * z = i * j in the context of this problem x + 30y = 120j x + 90y = 48j

4. perl

what is initial?

5. gingerkid101

the amount of grass that the field had initially

6. experimentX

yeah i figured out that.

7. perl

so you set them equal to each other because you assumed they ate all the grass

8. gingerkid101

I didn't assume, they said that they "cleared the field"

9. perl

errr, that was given

10. perl

so what exactly are you equating , what is your left side and right side

11. perl

total grass grown = total grass eaten (over the days )

12. gingerkid101

the initial grass plus the growth of the grass is equal to the grass that the cows ate in other words, when you subtract the total grass minus the grass eaten, it should equal zero, thus my equation.

13. perl

ok but thats 2 equations in 3 unknowns

14. perl

the grass is growing while they are eating

15. experimentX

grass growth must be some function of days

16. sritama

i started this way, 1 cow eats an ammount of grass each day is m the grass grows by the amount n each day and if the initial is a then we get a+90n=48*90m a+30n=120*30 m

17. perl

well you can subtract equation 1 from equation 2, that eliminates x the initial

18. sritama

by eliminating the initial i will find the relation between m and n

19. perl

how did you get grass growth rate is 90 and 30 ?

20. perl

nevermind

21. perl

that was days

22. gingerkid101

Yeah sritama I was actually just about to correct my equations to add the days on both sides.

23. sritama

yeah,i took the grass growth rate n/each day ... so it must b 90n and 30n

24. sritama

oh ok gingerkid 101

25. perl

ginger can you redo your equation

26. perl

i like your equation, its very logical :)

27. gingerkid101

x + 30y = 120 * 30 * j x + 90y = 48 * 90 * j where x is in grass, y is in grass per day j is in grass per cow per day

28. perl

yeah thats a little odd, grass per cow per day

29. perl

how does that come out in units, grass/ (cow/day) or

30. sritama

probably it means the amount of grass taken by each cow per day

31. gingerkid101

grass/(cow * day) or grass/cow/day it's all the same.

32. gingerkid101

in other words, if you have a certain number of cows, you can multiply it by that number to get 30 cows * 1 grass/cowday = 30 grass/day is resulting from having 30 cows.

33. gingerkid101

anyway the equations fully played out are... x + 30y = 3600j x + 90y = 4320j that's what I'm at right now.

34. perl

ok now its linear algebra, one moment

35. perl

x + 30y - 3600j=0 x + 90y- 4320j=0

36. perl

I get x = 3240 j y = 12j j = j

37. perl

actually we know we want 16 , so we have another equation

38. perl

x + 30y = 3600j x + 90y = 4320j x + 16 y = 16 i * j

39. gingerkid101

However, we also know that 8 additional cows are needed to balance the original proportion, which can possibly give us a 3rd equation? 8j - 60y = 0 or else the proportion would be even. therefore, we now have a 3 part system 8j - 60y = 0 x + 30y - 3600j=0 x + 90y- 4320j=0

40. perl

how did you get 8 additional cows?

41. sritama

yeah,thats my question too

42. gingerkid101

8 additional to 1/3 of 120. maybe I'm just talking out of my retricenow haha

43. gingerkid101

I don't even know. meh, I never had to do anything like this and I took up to calc 3 haha

44. perl

well this is a solid start

45. perl

i got 12j = y , so 12 cows it takes to eat the grass grown in 1 day