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    Mathematics - Calculus - Maximizing Area

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    A developer wants to enclose a rectangular grassy lot that borders a city street for parking. If the developer has 640 feet of fencing and does not fence the side along the street, what is the largest area that can be enclosed?

    © BrainMass Inc. brainmass.com December 24, 2021, 7:49 pm ad1c9bdddf
    https://brainmass.com/math/basic-calculus/maximizing-area-fence-221913

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    Let the length of the lot be x and width be y. Then total length of fence required =
    x + y + y = x + 2y
    Therefore, x + 2y = 640  x = 640 - 2y  (1). Also, differentiating wrt x, we get,
    1 = -2.(dy/dx)  dy/dx = -1/2  (2)
    Area of the enclosure A = xy = (640 - 2y)y = 640y - 2y^2 [From (1)]
    Differentiating wrt x, we get dA/dx = 640.(dy/dx) - 4y.(dy/dx) = 0
     640(-1/2) - 4y(-1/2) = 0  y = 160. So, x = 640 - 2(160) = 320.

    The desired dimensions are length = 320 ft and width = 160 ft.
    Maximum area that can be enclosed = 320 x 160 = 51200 sq.ft.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 7:49 pm ad1c9bdddf>
    https://brainmass.com/math/basic-calculus/maximizing-area-fence-221913

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