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Thread: Optimization problem, fencing

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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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    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

    $\displaystyle A = xy$ and $\displaystyle P = 3x + 2y$, 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 $\displaystyle A = 1.5\cdot 10^6 \,\textrm{ft}^2$.

    So $\displaystyle xy = 1.5\cdot 10^6$

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


    Therefore

    $\displaystyle P = 3x + 2y$

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

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

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



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

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


    This is minimised when the derivative is 0, so

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

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

    $\displaystyle x^2 - 10^6 = 0$

    $\displaystyle x^2 = 10^6$

    $\displaystyle x = 10^3$.


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

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

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

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

    $\displaystyle = \frac{6}{ 10^3}$

    $\displaystyle > 0$.

    So this is definitely the minimum.


    When $\displaystyle x = 10^3$

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

    $\displaystyle = 1.5\cdot 10^3$.



    So the dimensions are:

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