[Guest]

7 cows can eat 3 acres of grass of a certain grassland in 30 days, and 5 cows can eat 2 acres of grass of that grassland in 10 days.

The grass grows at a constant rate and each cow eats at a constant rate. The length of the grass before the cows start grazing is constant.

Determine the time it will take for 16 cows to eat 7 acres of grass of the grassland.

[Guest]

7 cows can eat 3 acres of grass of a certain grassland in 30 days, and 5 cows can eat 2 acres of grass of that grassland in 10 days.

The grass grows at a constant rate and each cow eats at a constant rate. The length of the grass before the cows start grazing is constant.

Determine the time it will take for 16 cows to eat 7 acres of grass of the grassland.

umm is it not 30 days?

[Prof. Templeton]

7cows * 30days = 3acres * i(initial amount of grass) + 3acres * 30days * g(grass growth)

and...

5cows * 10days = 2acres * i + 2acres * 10days * g

so...

210=3i + 90g

50=2i + 20g

6i + 180g - 420 = 6i + 60g - 150

120g = 270

g = 9/4

i = 5/2

16cows * n(number of days) = 7acres * 5/2 + 7acres * n * 9/4

16n = 35/2 + 63n/4

n = 70

So the answer is 70 days.

[Guest]

7 cows can eat 3 acres of grass of a certain grassland in 30 days, and 5 cows can eat 2 acres of grass of that grassland in 10 days.

The grass grows at a constant rate and each cow eats at a constant rate. The length of the grass before the cows start grazing is constant.

Determine the time it will take for 16 cows to eat 7 acres of grass of the grassland.

Well I have some trouble understanding this one.

Is the groupl of 7 cows and 5 cows different because given that 5 cows can consume 2 acres in 10 days means they can consume 6 acres in 30 days. This would indicate that the cow consumption rate is not the same for all cows. So I don't think I have enough information to solve this one, but I am new to this board so maybe it's just me.

[Guest]

7cows * 30days = 3acres * i(initial amount of grass) + 3acres * 30days * g(grass growth)

and...

5cows * 10days = 2acres * i + 2acres * 10days * g

so...

210=3i + 90g

50=2i + 20g

6i + 180g - 420 = 6i + 60g - 150

120g = 270

g = 9/4

i = 5/2

16cows * n(number of days) = 7acres * 5/2 + 7acres * n * 9/4

16n = 35/2 + 63n/4

n = 70

So the answer is 70 days.

while i think you are very intellegent i think your missing a very basic concept....

but i may be way off track myself

Im ashamed to even try to show my math but i believe the answer is 28 days.

Your use of the constant was great but i think not nessesary.

If you add twice the cows as given in the first equation...

ea: 7 cows eat 3 acres in 30 days

then by reason it would take:

14 cows to eat 6 acres in 30 days

thus if you add 2 more cows wouldnt it take less time to consume 6 acres and adding one acre to their path will not increase their abbility to consume that much.

So I believe your aproach is flawed because the answer is too far off from what common sence says they can eat.

and by your calculations if they limmeted their eating to what you say they will starve to death.

just my oppinion.... not worth allot.

[Guest]

I believe the wording is a bit confusing. The cows will eat at the same speed regardless of how fast the grass grows. If you replace "eat" with "clear" then it all works out.

Plus, the rate of grass growth has to be alarmingly fast for the initial conditions to be true.

[Guest]

The author said that each cow eats at a constant rate, not that they eat at the same rate.

If we analyze by groups...

Group 1

7 Cows eat 3 acres in 30 days

This group eats 1/10 acre per day.

Group 2

5 cows eat 2 acres in 10 days

This group eats 1/5 acre per day.

If we put them together, the groups eat 3/10 of an acre per day.

So, these 12 cows could eat 7 acres in 23 and 1/3 days.

The problem is, we must add 4 cows. Do these eat at the rate of group 1 or group 2, or at an entirely different rate? Should we average?

If we presume the additional 4 cows can eat exactly 1/3 of what the 12 cows are eating, then we have 4/10 of an acre being consumed daily. 7 acres would take 17 1/2 days.

That is my best guess anyhow.

Edited March 19, 2009 by BTH

[Prof. Templeton]

while i think you are very intellegent i think your missing a very basic concept....

but i may be way off track myself

Im ashamed to even try to show my math but i believe the answer is 28 days.

Your use of the constant was great but i think not nessesary.

If you add twice the cows as given in the first equation...

ea: 7 cows eat 3 acres in 30 days

then by reason it would take:

14 cows to eat 6 acres in 30 days

thus if you add 2 more cows wouldnt it take less time to consume 6 acres and adding one acre to their path will not increase their abbility to consume that much.

So I believe your aproach is flawed because the answer is too far off from what common sence says they can eat.

and by your calculations if they limmeted their eating to what you say they will starve to death.

just my oppinion.... not worth allot.

7cows * 30days = 3acres * i(initial amount of grass) + 3acres * 30days * g(grass growth)

and...

5cows * 10days = 2acres * i + 2acres * 10days * g

so...

210=3i + 90g

50=2i + 20g

6i + 180g - 420 = 6i + 60g - 150

120g = 270

g = 9/4

i = 5/2

16cows * n(number of days) = 7acres * 5/2 + 7acres * n * 9/4

16n = 35/2 + 63n/4

n = 70

So the answer is 70 days.

14 cows could eat six acres in 30 days. Let's use my previous equation and values for i and g. so...

14d = 6 * 5/2 + 6 * d * 9/4

14d = 30/2 + 54d/4

14d - 13.5d = 15

.5d = 15

d = 30

so yes 14 cows could eat 6 acres in 30 days, but whats going on in the other acre where the other two cows are? Can they keep up with the grass growth? Let's see...

2d = 1 * 5/2 + 1 * d * 9/4

2d = 5/2 + 9d/4

2d - 2.25d = 2.5

-.25d = 2.5

d = -.625

Nope. Those two cows over there in the other acre by themselves have been eating grass slower then it's been growing and they have been doing it for 30 days now. When the other 14 cows finish their six acres and come over to help out the grass is now higher then when they started eating the six acres so it takes longer, a lot longer.

[Guest]

I am too rusty. You are absolutely correct about the growth that is happening! Good job professor.

[Guest]

7cows * 30days = 3acres * i(initial amount of grass) + 3acres * 30days * g(grass growth)

and...

5cows * 10days = 2acres * i + 2acres * 10days * g

so...

210=3i + 90g

50=2i + 20g

6i + 180g - 420 = 6i + 60g - 150

120g = 270

g = 9/4

i = 5/2

16cows * n(number of days) = 7acres * 5/2 + 7acres * n * 9/4

16n = 35/2 + 63n/4

n = 70

So the answer is 70 days.

I just finished this on my own using a different method. I got different constants values, but my final answer still came out to...

70 days.

[Prof. Templeton]

It take 4-3/4 cows to eat one acre in one day. Moooo

Edited March 19, 2009 by Prof. Templeton

[bonanova]

I agree with the Professor

Cows eat the grass while the grass grows. There are two competing rates.

1. The number of acres/day a cow could eat if the grass did not grow at all: c [acres/day]
   
2. The inverse of the number of days a field would take to replenish itself after being eaten: g [/day].
   

Note that a big field grows back as fast as a little field; to get growth in acres/day, multiply g by the size of the field.

For the 3 acre field: Start with 3 acres, add 90g and subtract 210c to get 0. [3g - 7c acres/day] x 30 days

For the 2 acre field: Start with 2 acres, add 20g and subtract 50c to get 0. [2g - 5c acres/day] x 10 days

3 + 90g - 210c = 0

2 + 20g - 50c = 0

c = .4 acres/day/cow

g = .9 /day

For the 7 acre field

7 + 7Dg - 16Dc = 0 where D is the number of days 16 cows require to eat 7 acres of growing grass.

D = 70.

[bonanova]

It take 4-3/4 cows to eat one acre in one day. Moooo

Edit: Agree again. (1 + g) / c

[Guest]

14 cows could eat six acres in 30 days. Let's use my previous equation and values for i and g. so...

14d = 6 * 5/2 + 6 * d * 9/4

14d = 30/2 + 54d/4

14d - 13.5d = 15

.5d = 15

d = 30

so yes 14 cows could eat 6 acres in 30 days, but whats going on in the other acre where the other two cows are? Can they keep up with the grass growth? Let's see...

2d = 1 * 5/2 + 1 * d * 9/4

2d = 5/2 + 9d/4

2d - 2.25d = 2.5

-.25d = 2.5

d = -.625

Nope. Those two cows over there in the other acre by themselves have been eating grass slower then it's been growing and they have been doing it for 30 days now. When the other 14 cows finish their six acres and come over to help out the grass is now higher then when they started eating the six acres so it takes longer, a lot longer.

I bow down to your superior intelegence. And I now agree with your origional conclusion.

[Prof. Templeton]

I bow down to your superior intelegence. And I now agree with your origional conclusion.

[Guest]

So the way I approached the problem was different than the other posts...

If you try to figure out this problem by assuming the grassland is of very large size (i.e. >> 7 acre),

it becomes a bit more difficult.

With the area no longer finite, the idea of simultaneous equations flies out the window (having an initial area i=2,3, or 7 is no longer valid since i is very large~infinity).

However, some keen assumptions can be made...

1. Since the cows are no longer confined to just eating in the 7 acre area, the area of eaten grass simplifies to a function of time...

#acres eaten (at time t) = #acres grass eaten by cows (at time t) - #acres grass regrown (at time t)

also note #acres eaten (at t=0) = 0 (grass is at a constant level at start)

2. The problem stated that the cows eat at a constant rate and the grass grows at a constant rate so...

-#acres grass eaten (at time t) = C1*t (C1 being the amount of grass 1 cow can consume in acres/day * #cows)

-there must be some time (call it x) where the grass that has been eaten grows to a point where it is "regrown".

Because the grass grows at a constant rate, this time is constant for any blade in the grasslands.

3. From the start of grazing (t=0) to t=x, the first blade of grass hasn't reached the "regrown" state so the equation simplifies to...

#acres eaten (at time t) = C1*t

4. When time t=x, the first grass eaten becomes "regrown" and the right hand term comes into effect. However...

-The rate at which the grass is now accumulating is equal to the rate at which the cows are eating.

(i.e. if you cut your grass at say 1 mph from left to right and let it grow to 6", the grass height

would reach 6" at the left side first and each adjacent blade would reach 6" moving from left to right at 1 mph)

-This causes the #acres eaten (at t>x) to be a constant so

#acres grass eaten by cows (at t>x) = #acres grass regrown (at t>x)

The final equation for acres of grass eaten becomes a piecewise function where

for 0<t<x #acres eaten = C1*t

for t>or=x #acres eaten = C1*x (the maximum possible area that is eaten at any time given #cows)

Now we are left with the task of finding C1 and x.

Given the two data points, we can infer 3 possible cases

1. Both points fall at t>x

2. One point is at t<x and one is at t>x or t=x (best case)

3. Both fall at t<x

If the 5 cows eating 2 acres in 10 days is at t<x, C1 = 2acre/5cows/10days * #cows (.04 acre/cows days... if there are 16 cows, C1= .64 acre/day)

If the 7 cows eating 3 acres in 30 days is at t>or=x, x = 3acre/7cows/(.04 acre/cows days) = 10.71 days (notice how this number is independent of the #cows)

How do we determine Which case is true?

If case 1 is true, then the #acres eaten/cow would be the same for both points. Since 2/5 does not equal 3/7 (.4 and .429, respectively)

Case 1 is False

If case 3 were true, then the C1's generated from both data points (on a acre/cows days basis) should be equal (same slope). 2/5/10= .04 and 3/7/30= .014

Case 3 is False

Case 2 must be accepted (especially since x=10.71 days > 10 days)

So C1=.04 acre/cows days * #cows and x=10.71 days

With this data, we can finally calculate the time it takes 16 cows to reach 7 acres.

First lets see the maximum area possible (t=10.73 days)

Max #acres eaten = .64 acre/day * 10.71 days = 6.86 acres so it is impossible for 16 cows to reach 7 acres.

The minimum #cows needed is 17 with a max area of 7.29 acres

This is kind of a mute point considering the answer is that there is no answer but I solved it this way before looking at the thread.

Kind of a warped way of looking at the problem though.

Edited March 19, 2009 by Oobernater