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Posted by on Monday, February 18, 2013 at 9:26am.

A rectengular area is to be enclosed on three sides by fencing, an existing hedge forming the fourth side. Find the greatest area of the rectangle and its dimensions if 140mof fencing is available.

  • Math - , Monday, February 18, 2013 at 9:53am

    let the length parallel to the hedge be y
    let the width be x

    so y + 2x = 140
    y = 140-2x

    area = xy
    = x(140-2x) = -2x^2 + 140x

    if you know Calculus:
    -4x + 140 = 0
    x = 35
    then y = 140 - 70 = 70

    max area = 35(70) = 2450

    If no Calculus:
    complete the square to find the vertex

    A = -2x^2 + 140x
    = -2(x^2 - 70x)
    = -2(x^2 - 70x + 1225 - 1225)
    = -2( (x-35)^2 - 1225)
    = -2(x-35)^2 + 2450
    so the vertex is (35,2450)
    giving us a max of 2450 when x = 35

  • Math - , Thursday, May 30, 2013 at 9:20pm

    IDK but I'm not in eight grade, YET!! Comment, share, and love My pro in Facebook. Find me as Jessi Vertiz

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