Calculus 1

If you have 280 meters of fencing and want to enclose a rectangular area up against a long, straight wall, what is the largest area you can enclose?

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  1. Let the perimeter be 2w + 280-2w = 280

    A = w * (280-2w)
    A = 280w - 2w^2
    dA/dw = 280 - 4w

    A is max when dA/dw = 0

    Thus, w=70

    The dimensions are 70x140 and the area is 9800

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  2. Area A=x.y
    The perimeter of a rectangle is 2x+y= 280 solving for y give you,m y=280-2x, y is greater to zero.
    A=x*y substituting y in the area equation give you
    A=x(280-2x.)=280x-2x^2
    The area is maximized when A'=0
    so first find the derivative of A
    A'=280-4x. A=0 when 280-4x=0 solve for x. x=280/4=70 so x=70
    you subtitue x into the y equation which is y=280-2x so y = 280-2(70)=280-140=140
    so y=140
    Finally the dimensions are x=70, and y=140 so the area is 70*140=9800m^2

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