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a farmer with 10,000 meters of fencing wants to enclose a rectangular field and then divide it into two plots with a fence parallel to one of the sides. What is the largest area that can be enclose?

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2 answers
  1. 3 w + 2 L = 10,000
    so
    w = (10,000 -2 L)/3

    A = w L = (10,000/3)L -(2/3)L^2
    that is a parabola (quadratic)
    we could find the vertex or we could use calculus
    first find the vertex
    A is 0 when L = 0
    A is 0 when (2/3)L =10,000/3
    or when L = 5000
    so vertex is halfway between
    L = 2500
    then
    w = (10,000 - 5,000) /3 = 5,000/3 = 1667

    Now with calculus
    dA /dL = w dL/dL + l dw/dL = w + L dw/dL
    when dA/dL = 0 we have an extreme
    w = -L dw/dL
    but w = 10,000/3 - (2/3) L
    so
    dw/dL = -2/3
    so
    w = (2/3) L
    so
    2 w = 10,000/3
    w = 5000/3
    then L = 10,000/4 = 2,500
    which we knew from the parabola anyway

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  2. a farmers wants fence 250 h/a of land,
    how much meters are in there?

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