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Math Help - Quadratic Problem

  1. #1
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    Quadratic Problem

    You have $3000 with which to build a rectangular enclosure with fencing. The fencing material costs $20 per meter. You also want to have two partitions across the width of the enclosure, so that there will be three separated spaces in the enclosure. The material for the partitions costs $15 per meter. What is the maximum area you can achieve for the enclosure? (Round your answer to the nearest whole number.)

    I am confused with what I would have to do with the $20 and $15 constraints in this problem?
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  2. #2
    MHF Contributor
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    Re: Quadratic Problem

    Hey UWM120.

    Do you need fencing surrounding the whole thing or can you use the partitions to separate the outside from the inside?

    If you need fencing to separate the outside from the inside then you only need to maximize the area given the fencing and this for a rectangular region will always result in a square (if you want to minimize the cost).
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  3. #3
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    Re: Quadratic Problem

    Hello, UWM120!

    You have $3000 with which to build a rectangular enclosure with fencing.
    The fencing material costs $20 per meter.
    You also want to have two partitions across the width of the enclosure,
    so that there will be three separated spaces in the enclosure.
    The material for the partitions costs $15 per meter.
    What is the maximum area you can achieve for the enclosure?
    (Round your answer to the nearest whole number.)

    Code:
          : . . .  x  . . . :
        - *-----*-----*-----* -
        . |     |     |     | .
        . |     |     |     | .
        y |    y|     |y    | y
        . |     |     |     | .
        . |     |     |     | .
        - *-----*-----*-----* -
          : . . .  x  . . . :
    There are two x-meter lengths at $20 per meter.
    There are two y-meter widths at $20 per meter.
    . . Their cost is 40x + 40y dollars.

    There are two y-meter partitions at $15 per meter.
    . . Their cost is: 30y dollars.


    The total cost is: 40x + 70y dollars which will equal $3000.
    . . 40x + 70y \:=\:3000 \quad\Rightarrow\quad y \:=\:\frac{300-4x}{7} .[1]

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

    Substitute [1] into [2]: . A \;=\;x\left(\frac{300-4x}{7}\right) \;=\;\tfrac{1}{7}(300x-4x^2)

    And that is the function we must maximize.

    Go for it!
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