## how many horses can be fed on 3rd field for eighteen weeks?

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little_dragon
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### how many horses can be fed on 3rd field for eighteen weeks?

Suppose there are three pastures with grass of identical height, density, and growth rate. The first is ten acres and can feed twenty-one horses for nine weeks. The second is three and a third acres and can feed twelve horses for four weeks. The third is twenty-four acres. How many horses can be fed on this third field for eighteen weeks?

In the first field, the ten acres has (21)(9) = 189 horse-weeks of food. Thats 189/10 = 18.9 horse-weeks / acre.

In the second field, the 3 1/3 acres has (12)(4) = 48 horse-weeks of food. thats 48 / (10/3) = 14.4 horse-weeks / acre.

The rates dont match, so what do I do now?

stapel_eliz
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This is actually somewhat complex. You need to take into account the fact that, while the grass is being eaten, it is also growing. At some point, the fields are grazed to the point of uselessness, but it isn't just a matter of the grass being eaten; you also have to account for regrowth.

From the first field, we get that 21 horses take 9 weeks to graze the 10 acres down to the ground. During that time, the grass is growing at some rate G cubic centimeters per week per acre. (It's "cubic" because we're taking the area of an acre and multiplying by Since the fields are assumed to start from the same conditions, this means that we have 10 acres that start at some volume, based on the original height, of H cc/acre of grass. The horses eat this at some rate E cc per horse per week. So, after nine weeks at this rate, we have

. . . . .(10 acres)(H cc/acre) + (9 weeks)[(10 acres)(G cc/acre-week)
. . . . . . . . - (21 horses)(E cc/horse-week)] = 10H + 90G - 189E = 0

In the same way, the second field gives us:

. . . . .(10/3 acres)(H cc/acre) + (4 weeks)[(10/3 acres)(G cc/acre-week)
. . . . . . . .- (12 horses)(E cc/horse-week)] = (10/3)H + (40/3)G - 48E = 0

Then what? We have two equations, but with three variables:

. . . . .10H + 90G - 189E = 0
. . . . .(10/3)H + (40/3)G - 48E = 0

We know that we'll have 24 acres for the third field, and horses will be grazing there for 18 weeks. So we know that the third equation will be:

. . . . .24H + 432G - (x horses)(18E) = 0

If we solve the first two equations for "10H=", we can get a ratio of values:

. . . . .10H = 198E - 90G
. . . . .10H = 144E - 40G

Then:

. . . . .198E - 90G = 144E - 40G

. . . . .54E = 50G

. . . . .27E = 25G

. . . . .E = (25/27)G

Then:

. . . . .10H = (144)(25/27)G - 40G

. . . . .10H = (400/3)G - (120/3)G = (280/3)G

. . . . .H = (28/3)G

Does this help? Let's see:

. . . . .24H + 432G - (x horses)(18E) = 0

. . . . .24H + 432G - 18x(25/27)G = 0

. . . . .(24)(28/3)G + 432G - (18x)(50/3)G = 0

. . . . .200G + 432G - 300Gx = 0

. . . . .632G - 300Gx = 0

You can solve this for "x=", since the two Gs will cancel out.

Check my math, though; because it seems reasonable that you should end up with a whole number. I'll be back later to proof the above....

little_dragon
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### Re: how many horses can be fed on 3rd field for eighteen weeks?

I think "24H + 432G - 18x(25/27)G = 0" goes to "24(28/3)G+432G-2x(25/3)G=0" and then "224G+432G-(50/3)xG=0". Then x=(656G)/(50G/3)=39.36. But that's still not a whole number. Maybe i should round?

i also tried it this way: the 1st field has 10H+90G of grass that the horses eat in 9 wks, so they each eat (10H+90G)/(189) /wk. the 2nd field has them eating (10H/3+40G/3)/48 /wk. these are the same, so H=12G. then the 3rd field gives (24H+432G)/18x /wk. this gives x=36.

i can't find the mistke doing it the other way but this way works. thanks for the help.

this was way more complicated than i'd thought it was going to be.