Cow, sheep and horse field problem

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pazzy78

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Someone sent me this :
A horse, cow, and a sheep get enough feeding out of one field for 20 days
If the cow and horse get 25 days out of it, the cow and sheep 33 and a third days, the horse and sheep 50 days, how long will the cow alone get out of it
And the sheep alone
And the horse alone??
Click to expand...

Now let C = total daily amount a cow eats
S = total daily amount a sheep eats
H = total daily amount a horse eats

X = total volume of food/grain in field.

So these equations make perfect sense to me.

20C + 20S + 20H = X
25H + 25C = X
33.3333C + 33.3333S = X
50H + 50S = X

But immediately I see I am wrong as I get 33.3333d for a cow to use up his supply and we know a cow + sheep will take 33.333 days to eat the whole field.
 
hmm The more I think about this, the more I am thinking of rates and a calculas problem, so let's call X above Y ... just for simplicity sake I don't like dX on the bottom :)

something like dY/dC = rate of change of the vol. of field with respect to cows eating ? etc ?

or am I over complicating it ?


edit maybe not ...
 
pazzy78 said:
Someone sent me this :


Now let C = total daily amount a cow eats
S = total daily amount a sheep eats
H = total daily amount a horse eats

X = total volume of food/grain in field.

So these equations make perfect sense to me.

20C + 20S + 20H = X
25H + 25C = X
33.3333C + 33.3333S = X
50H + 50S = X

But immediately I see I am wrong as I get 33.3333d for a cow to use up his supply and we know a cow + sheep will take 33.333 days to eat the whole field.
Click to expand...
It appears the problem is wrong. I think if you finish solving (I did it a little differently), you'll find that X = 0!

(I would use fractions rather than decimals in the third equation; and I initially replaced your X with 100 (meaning 100% of the field), and found the equations to be inconsistent.)
pazzy78 said:
hmm The more I think about this, the more I am thinking of rates and a calculas problem, so let's call X above Y ... just for simplicity sake I don't like dX on the bottom :)

something like dY/dC = rate of change of the vol. of field with respect to cows eating ? etc ?

or am I over complicating it ?


edit maybe not ...
Click to expand...
No, this isn't a rate of change problem.
 
Dr.Peterson said:
It appears the problem is wrong. I think if you finish solving (I did it a little differently), you'll find that X = 0!
Click to expand...
When I saw 4 equations with 3 unknowns my first question was: are they consistent? Seems they aren't.
 
omg .. I just found this

not the same numbers but the same issue - inconsistent ... doesn't take growth into account ... I mean ....


Feck off!!

Ridicolous, I wasted to much time on that...

Something like growth should be included ... like "Assume the grass grows at a constant rate" etc...
 
Adding the growth constant into it didn't help me much, they just cancelled out and still faced the same inconsistency...
 
pazzy78 said:
omg .. I just found this

not the same numbers but the same issue - inconsistent ... doesn't take growth into account ... I mean ....

Something like growth should be included ... like "Assume the grass grows at a constant rate" etc...
Click to expand...
Now that I think of it, I have seen such a problem before, where growth was specifically mentioned, making it a rate of change problem, as I said it wasn't. (But it isn't a calculus problem in any case.)

I wouldn't have thought of that in this problem, because, as you say, nothing in the wording points toward that. The words "graze the field bare" in the alternative problem at least hints at something other than the steady state that most artificial problems assume. But maybe if I raised animals, I'd have known from experience that it isn't steady state! An example where problem-domain knowledge is required to interpret it.
pazzy78 said:
Adding the growth constant into it didn't help me much, they just cancelled out and still faced the same inconsistency...
Click to expand...
Can you show your work again? I'll see what I get... yes, I do get a neat solution.
 
20C + 20S + 20H = X
25H + 25C = X
33.333C + 33.333S = X
50H + 50S = X

so let g = constant growth..

20C + 20S + 20H = X + 20g
25H + 25C = X + 25g
33.333C + 33.333S = X + 33.333g
50H + 50S = X + 50g


Multiply 1st eq by 2.5

50C + 50S + 50H = 2.5X + 50g (subtract 4th eq)
- - 50S - 50H = -X -50g
---------------------------------

50C = 1.5X

C = 0.03X

So a Cow eats 3% of the field per day ....

1 cow would take 33 1/3 days to finish the field - but we are told a cow and a sheep grazes it bare in 33 1/3 days...



:mad::mad::mad::mad::mad:
 
Person who originally sent it to me told me there is no growth aspect to it ... so we still get the same inconsistency without the g ...
 
pazzy78 said:
So a Cow eats 3% of the field per day ....

1 cow would take 33 1/3 days to finish the field - but we are told a cow and a sheep grazes it bare in 33 1/3 days...
Click to expand...
But during those 33 1/3 days, the grass is growing ...

Think about how the author got his answer:
c = G/3, h = G/4, s = G/6, and g = G/4.
Click to expand...
So, assuming that the grass grows at a constant rate per day, the cow alone can graze the field bare in 12 days,
but the horse can only keep up with the new growth and will never graze the field bare at all, while the sheep will be
overwhelmed because the grass will be growing half as fast again as the sheep can eat it
Click to expand...
He took into account both c and g to get the 12.

pazzy78 said:
Person who originally sent it to me told me there is no growth aspect to it ... so we still get the same inconsistency without the g ...
Click to expand...
How do you know he was right??

To clarify: We get the inconsistency without the growth rate; we need it to make it consistent.
 
Last edited:
So sorry, he sent me the complete problem ... there is growth , it says the grass is cut to 2 inches in height and grows at a uniform rate ...

Again sorry about this , I wasted a lot of time in this too , sorry to share an incomplete problem
 
How does [for the other problem] he get 12 days for the cow ? The cow eats ⅓ of the field in a day
... I understand the field grows a ¼of it back per day ... so in those 3 days only an extra ¾ of the field would be replaced ?

How.does he arrive at 12 ?
 
I've got it!!!

Cow 50days
Sheep - never ends, his eating rate = the growth rate
Horse - 100 days

20240409_015944.jpg
 
The details of my work are a little different, but you've got it.

I got (using your variables) C = 0.03G, S = 0.01G, H = 0.02G, g = 0.01G.

Taking growth into account, with the cow alone the net loss per day is 0.03G-0.01G = 0.02G, so G is all used up in 1/0.02 = 50 days.

Similarly, the sheep alone matches the growth rate (net rate 0), so it never has to stop; and the horse alone gives a net rate of 0.02G - 0.01G - 0.01G, so it uses up all of G in 100 days.
 
Yes, a pity the .03G value was putting me off ... as it was giving 33.333... days for a cow - but not taking growth into account...
 
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