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20 January, 02:04

A rancher has 800m of fencing to enclose a rectangular cattle pen along a riverbank. No fencing is required along the riverbank. Find the dimensions that would maximize the area for the cattle to graze. Also, what is the maximum area?

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  1. 20 January, 02:05
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    Let the dimensions be L and W. Then P = perimeter = 800 m = 2W + L.

    Constraint:

    We want to maximize the area. This area is L*W. We can eliminate one variable or the other by solving 800=2W + L for either. L = 800 - 2W. Then

    the area is A (W) = (800 - 2W) * W.

    Differentiate this with respect to W: A ' (W) = (800 - 2W) (1) + W (-2), or

    800 - 2W - 2W = 800 - 4W

    Set this derivative = to 0 and solve for W: W = 800/4, or W = 200. Then L = 800 - 2W becomes L = 800 - 2 (200) = 400.

    Thus, W = 200 m and L = 400. From this data we get max area = (200) (400) m^2, or 80000 m^2.
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