Grazing cows

[BMAD]

Freddie the farmer has a paddock that he uses to graze his stock of cows. Each cow eats the same amount of grass each day, regardless of how many other cows are in the paddock and irrespective of the amount of grass left in the paddock. In an experiment, Freddie puts 6 cows into the paddock and he finds out it takes 3 days for them to eat all the grass. These 6 cows are then taken out of the paddock to allow the grass to grow back.

After the grass has been allowed to grow back to the original amount, Freddie then puts 3 cows into the paddock. This time he finds that it takes 7 days for the 3 cows to eat all the grass in the paddock. Freddie is puzzled that the cows took this long and consults a mathematician.

Freddie said "Geez mate, I dunno why me cows took that long to eat me paddock."

Marvin the mathematician replies "Well Freddie, tell me what assumptions you made."

Freddie replies "Well mate, maybe me cows got sick or somethin', cos I reckon that me 3 cows should have taken 6 days to eat me paddock, not 7 days! "

Marvin Replies, "I doubt that very much Freddie!"

After a while Marvin does some calculations and reveals that Freddie had overlooked an important assumption. What was the assumption Freddie had overlooked? Using Marvin's assumption, how long would a single cow take to eat the same paddock?

[mattil]

Was the assumption that each cow eats the same amount as every other cow? If each cow eats a different amount of food then the time it takes for one cow to graze the paddock would depend on which particular cow was placed in the paddock.

[BMAD]

nope. that assumption was safely given.

[harey]

Well, Freddie overlooked that the grass continues to grow. Whether it is an assumption?

I rather fear that we have to assume that the same amount of grass grows every day, regardless of the stock. Maybe a farmer could confirm.

a - stock of grass
b - day's growth
c - quantity eaten by a cow a day

1) a+3*b=6*3*c
2) a+7*b=3*7*c
1)-2) 4b=3c

So one cow needs 3/4 of a day to eat one day's increase, leaving 1/4 of a day to eat the stock. If I did not miscalculate, 63/4 days.

Edited November 6, 2013 by harey

[BMAD]

Well, Freddie overlooked that the grass continues to grow. Whether it is an assumption?

I rather fear that we have to assume that the same amount of grass grows every day, regardless of the stock. Maybe a farmer could confirm.

a - stock of grass

b - day's growth

c - quantity eaten by a cow a day

1) a+3*b=6*3*c

2) a+7*b=3*7*c

1)-2) 4b=3c

So one cow needs 3/4 of a day to eat one day's increase, leaving 1/4 of a day to eat the stock. If I did not miscalculate, 63/4 days.

I agree that you found the 'missed assumption' by the farmer but i calculated a different answer. Someone please verify.

[plasmid]

I came up with the same answer as harey for the first part (where he shows his work

): 4b = 3c. I differ thereafter.

Defining N as the number of days that a single cow could be supported by the paddock:

a + Nb = Nc

a + 3b = 6 * 3 * c

Subtracting the two equations

(N-3)b = (N-18)c

(N-3) * 3*4*b = (N-18) * 4*3*c

(N-3) * 3 = (N-18) * 4

3N - 9 = 4N - 72

N = 63

Edited November 7, 2013 by plasmid

[harey]

The first part did not require much calculation . As usually, I made an arithmetic error. I tried with b=60, c=80 - plasmid is correct.

[BMAD]

I came up with the same answer as harey for the first part (where he shows his work

): 4b = 3c. I differ thereafter.

Defining N as the number of days that a single cow could be supported by the paddock:

a + Nb = Nc

a + 3b = 6 * 3 * c

Subtracting the two equations

(N-3)b = (N-18)c

(N-3) * 3*4*b = (N-18) * 4*3*c

(N-3) * 3 = (N-18) * 4

3N - 9 = 4N - 72

N = 63

this is what i have as well.

[DeGe]

You assume a constant rate of growth of grass over the entire duration.

I'm not sure about this and pls correct me if I am wrong: If some grass is eaten on any given day, shouldn't the rate of growth of grass be slower the next day?

[BMAD]

You assume a constant rate of growth of grass over the entire duration.

I'm not sure about this and pls correct me if I am wrong: If some grass is eaten on any given day, shouldn't the rate of growth of grass be slower the next day?

i think the rate would be the same in terms of average growth

[OldGeoff]

Grazing Cows

Alas the answers suggested by BMAD and plasmid were both incorrect.

Let a = initial amount of grass, in kg say

b = quantity of growth, in kg per day

c = quantity consumed by one cow, in kg per day

k = number of cows and

N = number of days it takes to consume all available grass

Now we can write:

initial quantity of grass + growth over N days = quantity consumed by k cows in N days,

a + bN = kcN

Making k the subject we can determine the relationship between k and N:

k = a/c (1/N) + b/c where a, b and c are constants.

Thus the relationship is a hyperbola with asymptotes N = 0 and k = b/a.

The diagram above shows a hyperbola passing through the two fixed points, (6 cows, 3 days)

and (3 cows, 7 days), but there will be a family of such curves, depending on the ratio b/a, the

quantity of growth per day to the original quantity.

As for the question of how many days it would take one cow to consume the total feed, if we let

k = 1 and solve for N we obtain N = a /(c -b)

Thus the time taken for one cow to eat all the feed depends on a, the original quantity of feed, and (c-b), the difference between c, the quantity eaten per day, and b, the growth per day.

If c > b N will be some finite number of days, if c ≤ b the cow will never catch up with the growth.

[OldGeoff]

Diagram is missing from previous post. I don't seem to be able to post it.