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Math Help - Linear algebra question

  1. #1
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    Linear algebra question

    Hi. I thing I'm missing something simple, but can anyone help me solve the following:

    A role of fence wire 1200m long is to be used to enclose and seperate two identical adjoining rectangular paddocks (which share one side-width).

    Find the dimensions of the enclosure, so the total area is maximised.

    Thanks in advance for the help.
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  2. #2
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    Re: Linear algebra question

    A = WL

    2W + 3L = 1200

    W = 600 - 3L/2

    A(L) = (600 - 3L/2)L = 600L - 3L2/2

    take the derivative of A with respect to L (dA/dL) and set it equal to 0, then solve for L.
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  3. #3
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    Re: Linear algebra question

    Hello, elmidge!

    Did you make a sketch?


    A roll of fence wire 1200m long is to be used to enclose and seperate
    two identical adjoining rectangular paddocks (which share one side-width).

    Find the dimensions of the enclosure so the total area is maximised.

    Let x = total length of the paddocks.
    Let y = width of the paddocks.
    Code:
          : - - - - - x - - - - - :
          *-----------*-----------*
          |           |           |
          |           |           |
         y|           |y          |y
          |           |           |
          |           |           |
          *-----------*-----------*
          : - - - - - x - - - - - :
    We have: . 2x + 3y \:=\:1200 \quad\Rightarrow\quad y \:=\:400 - \tfrac{2}{3}x .[1]

    The area is: . A \:=\:x\cdot y

    Substitute [1]: . A \:=\:x\left(400-\tfrac{2}{3}x\right) \quad\Rightarrow\quad A \:=\:-\tfrac{2}{3}x^2 + 400x

    This is a down-opening parabola: \cap
    Its maximum is at its vertex: . x \:=\:\tfrac{\text{-}b}{2a}

    Hence: . x \:=\:\frac{\text{-}400}{2(\text{-}\frac{2}{3})} \quad\Rightarrow\quad \boxed{x\:=\:300\text{ m}}

    Substitute into [1]: . y \:=\:400 - \tfrac{2}{3}(300) \quad\Rightarrow\quad \boxed{y \:=\:200\text{ m}}
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