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Math Help - Optimization problem, fencing

  1. #1
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    Optimization problem, fencing

    11. A farmer wants to fence an area of 1.5 million square feet in a rectangular field and then divide it in half with a fence parallel to one sides of the rectangle. how can he do this so as to minimize the cost of the fence.

    i got this problem completely wrong, but here's what I did,
    area = 1500000
    area = xy = 1,500,000
    length of fence = perimeter = 3x+4y
    then
    xy=1,500,000
    y=\frac{1,500,000}{x}
    3x+4\frac{1,500,000}{x}
    f'(x) = 3+\frac{-6,000,000}{x^2}
    = \frac{3x^2-6,000,000}{x^2}
    = \frac{3(x^2-2,000,000)}{x^2}
    x=1414
    then... i got completely lost.
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  2. #2
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    Quote Originally Posted by dorkymichelle View Post
    11. A farmer wants to fence an area of 1.5 million square feet in a rectangular field and then divide it in half with a fence parallel to one sides of the rectangle. how can he do this so as to minimize the cost of the fence.

    i got this problem completely wrong, but here's what I did,
    area = 1500000
    area = xy = 1,500,000
    length of fence = perimeter = 3x+4y
    then
    xy=1,500,000
    y=\frac{1,500,000}{x}
    3x+4\frac{1,500,000}{x}
    f'(x) = 3+\frac{-6,000,000}{x^2}
    = \frac{3x^2-6,000,000}{x^2}
    = \frac{3(x^2-2,000,000)}{x^2}
    x=1414
    then... i got completely lost.
    Actually

    A = xy and P = 3x + 2y, since you only need one extra length to go down the middle. Since you want to minimise the cost of the fencing, you need to minimise the perimeter.


    You know A = 1.5\cdot 10^6 \,\textrm{ft}^2.

    So xy = 1.5\cdot 10^6

    y = \frac{1.5\cdot 10^6}{x}.


    Therefore

    P = 3x + 2y

     = 3x + 2\left(\frac{1.5\cdot 10^6}{x}\right)

     = 3x + \frac{3 \cdot 10^6}{x}

     = 3x + 3\cdot 10^6 x^{-1}.



    \frac{dP}{dx} = 3 - 3\cdot 10^6 x^{-2}

     = 3 - \frac{3\cdot 10^6}{x^2}.


    This is minimised when the derivative is 0, so

    3 - \frac{3\cdot 10^6}{x^2} = 0

    3x^2 - 3\cdot 10^6 = 0

    x^2 - 10^6 = 0

    x^2 = 10^6

    x = 10^3.


    To be a minimum, the second derivative needs to be positive at that point.

    \frac{d^2P}{dx^2} = 6\cdot 10^6 x^{-3}

     = \frac{6\cdot 10^6}{x^3}

     = \frac{6\cdot 10^6}{(10^3)^3} when x = 10^3

     = \frac{6}{ 10^3}

     > 0.

    So this is definitely the minimum.


    When x = 10^3

    y= \frac{1.5\cdot 10^6}{10^3}

     = 1.5\cdot 10^3.



    So the dimensions are:

    x = 10^3\,\textrm{ft} and y = 1.5\cdot 10^3\,\textrm{ft}, with x as the length used to divide the paddock in half.
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