algebra

A rancher has
4 comma 7004,700
feet of fencing available to enclose a rectangular area bordering a river. He wants to separate his cows and horses by dividing the enclosure into two equal areas. If no fencing is required along the​ river, find the length of the center partition that will yield the maximum area.

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  1. same as the others. Give it a try. period

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  2. A rancher has
    4,700 feet of fencing available to enclose a rectangular area bordering a river. He wants to separate his cows and horses by dividing the enclosure into two equal areas. If no fencing is required along the​ river, find the length of the center partition that will yield the maximum area.

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  3. You don't say whether the dividing fence is to be parallel to the river or not. If not, then let x be the length of the side parallel to the river. Then you have

    x+3y = 4700
    so, x = 4700-3y

    The area is
    a = xy = (4700-3y)y = 4700y - 3y^2
    The vertex of this parabola is at (2350/3,5522500/3) or (783.33,1.84*10^6)

    As usual, the maximum area is when the fence is divided equally among lengths and widths:
    one length of 2350
    3 widths of 2350/3

    Now, if the dividing fence op length x is parallel to the river, you have

    2x+2y = 4700
    a = x(2350-x) = 2350x-x^2
    and max area is at x=2350/2 and is 1.38*10^6

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