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Thread: Rectangular Swimming Pool

  1. #1
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    Rectangular Swimming Pool

    A rectangular swimming pool has length 30 m by 20 m. There is a deck of uniform width surrounding the pool. The area of the pool is the same as the area of the deck. Write a quadratic equation to model this situation and use it to determine the width of the deck.
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  2. #2
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    <----20+2x---->
    _____________
    |...................|...^
    |.....__20__... |....|
    |..x.|........ |...|...|
    |....|.........|30|...30+2x
    |....|_____ |...|....|
    |...................|....|
    |____________|... v


    Pool area = $\displaystyle 600m^2$

    Deck area = $\displaystyle (30 + 2x)(20 + 2x) = 600$

    $\displaystyle 4x^2 +100x - 600 = 0$

    $\displaystyle x^2 +25x - 150 = 0$

    $\displaystyle (x + 30)(x - 5) = 0$

    $\displaystyle x = 5$ (width can't be -30)

    width of deck = 20 + 2x = 20 + 10 = 30m.
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  3. #3
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    but.....

    Quote Originally Posted by nzmathman View Post
    <----20+2x---->
    _____________
    |...................|...^
    |.....__20__... |....|
    |..x.|........ |...|...|
    |....|.........|30|...30+2x
    |....|_____ |...|....|
    |...................|....|
    |____________|... v


    Pool area = $\displaystyle 600m^2$

    Deck area = $\displaystyle (30 + 2x)(20 + 2x) = 600$

    $\displaystyle 4x^2 +100x - 600 = 0$

    $\displaystyle x^2 +25x - 150 = 0$

    $\displaystyle (x + 30)(x - 5) = 0$

    $\displaystyle x = 5$ (width can't be -30)

    width of deck = 20 + 2x = 20 + 10 = 30m.
    I thank you for the reply.

    However, I am slightly lost using the FOIL method in this case.

    We have:

    (30 + 2x) (20 + 2x) = 600

    On the left side, I got this:

    4x^2 + 100x + 600 = 600

    I then subtracted 600 from both sides and got this:

    4x^2 + 100x = 0

    How did you get -600 on the left side of the equation?

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  4. #4
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    Sorry about that. It should say:

    $\displaystyle (30 + 2x)(20 + 2x) = 1200\,\,$ (because looking at the diagram $\displaystyle (30 + 2x)(20 + 2x)$ is the whole combined area of the deck and pool).

    $\displaystyle (30 + 2x)(20 + 2x) = 1200$

    $\displaystyle 4x^2 + 100x + 600 = 1200$

    $\displaystyle 4x^2 + 100x - 600 = 0$
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  5. #5
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    ok.............

    Quote Originally Posted by nzmathman View Post
    Sorry about that. It should say:

    $\displaystyle (30 + 2x)(20 + 2x) = 1200\,\,$ (because looking at the diagram $\displaystyle (30 + 2x)(20 + 2x)$ is the whole combined area of the deck and pool).

    $\displaystyle (30 + 2x)(20 + 2x) = 1200$

    $\displaystyle 4x^2 + 100x + 600 = 1200$

    $\displaystyle 4x^2 + 100x - 600 = 0$
    Now it makes sense.

    Thank you for your help.
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