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Math Help - Optimization problems #1

  1. #1
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    Optimization problems #1

    1. A farmer with 750 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens?

    How do I draw the diagram: one with shallow wide pens and the other with deep narrow pens. I assigned the variables which are: A = x * y but dont know if this is the correct variable. What is the correct equation to be differentiated? Is the domain (0,750)?
    Last edited by mr fantastic; October 5th 2009 at 03:29 AM. Reason: Re-titled post
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  2. #2
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    Hello, jsu03!

    1. A farmer with 750 ft of fencing wants to enclose a rectangular area
    and then divide it into four pens with fencing parallel to one side of the rectangle.
    What is the largest possible total area of the four pens?
    First, you need to visualize the problem . . . carefully.
    Code:
          : - - - - - x - - - - - :
          *-----*-----*-----*-----*
          |     |     |     |     |
          |     |     |     |     |
         y|    y|     |y    |y    |y
          |     |     |     |     |
          |     |     |     |     |
          *-----*-----*-----*-----*
          : - - - - - x - - - - - :

    The region is a rectangle with length x and width y.

    It is divided into 4 pens with three more fences of width y.

    The fencing is limited to 750 feet: . 2x+ 5y \:=\:750 \quad\Rightarrow\quad y \:=\:150 - \tfrac{2}{5}x .[1]


    The area of the rectangle is: . A \;=\;xy .[2]

    Substitute [1] into [2]: . A \;=\;x\left(150 - \tfrac{2}{5}x\right) \quad\Rightarrow\quad A \;=\;150x - \tfrac{2}{5}x^2

    Maximize: . A' \:=\:150-\tfrac{4}{5}x\:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:\tfrac{375}{2}}

    Substitute into [1]: . y \;=\;150-\tfrac{2}{5}\left(\tfrac{375}{2}\right) \quad\Rightarrow\quad\boxed{ y \:=\:75}


    The maximum area is: . A \;=\;\left(\tfrac{375}{2}\right)(75) \;=\;\boxed{14,\!062.5\text{ ft}^2}

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