A field is bound on one side by a river. A farmer wants to enclose the other three sides of the field with fence in order to create a rectangular plot of land for his cows. If the farmer has 400m of fence to work with, determine the maximum possible area of the field and the field's demensions.

(i'm on in grade 10, so I don't know calculus so don't use that)

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Anyway, we don' need no steeking calculus for this one.

If the side parallel to the river has length x, and the other side is y, then we have

x+2y = 400

The area a of the field is just xy, so

a = xy = (400-2y)y = 400y - 2y^2

This is just a parabola, opening downward, so it has a maximum value at the vertex, which occurs at y=400/4 = 100

so, if y=100, x=200 and the area is 20,000 m^2.

As is usual in this kind of problem, the maximum area is achieved when the fencing is divided equally among the lengths and widths. In this case 1 length of 200 and 2 widths of 100 each, making up the 400 total.

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posted by Steve

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