Math

A farmer has 54m of fencing with which to build two animal pens with a common side. one pen is rectangular; the other is square. if the area of the pens is to be maximized, what are their dimensions?

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  1. let the side of the square be x
    let the sides of the rectangle be x by y

    so 5x + 2y = 54
    y = (54-5x)/2

    area = x^2 + xy
    = x^2 + x(54-5x)/2
    = x^2 + 27x - (5/2)x^2

    d(area)/dx = 2x + 27 - 5x
    = 0 for a max of area
    3x = 27
    x = 9
    y = (54 - 45)/2 = 4.5
    the square is 9 by 9, and the rectangle is 9 by 4.5

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