a farmer has enough fencing to build 40 feet of fence. He wishes to build a rectangular pen nest to his barn wall forming one side of the pen. What dimension should he make the pen so as to enclose the greatest possibility area?

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  1. length = x
    width = y
    area = A = x y
    length of fencing = 2x+y = 40
    y = 40-2x

    A = x (40 - 2x)
    A = 40 x - 2 x^2

    maxmimum A , when dA/dx = 0
    0 = 40 - 4 x
    x = 10
    then y = 40 - 20 = 20
    area = A = 200

    alternatively look at parabola
    2 x^2 -40 x = -A
    x^2 - 20 x = -(A/2)
    x^2 - 20 x + 100 = -(A/2) + 100

    (X-10)^2 = -(1/2)(A+200)
    vertex at x = 10, A = 200

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  2. 40=2L+w

    area= LW=L(40-2L)

    Using calculus
    dArea/dL=0=40-2L + l(-2)

    L=10 W= 20 so the pen is 10x20,with the barn serving as a 20ft side.

    non calculus?
    graph area vs L. Where is it maximum?

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