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Math Help - equation set up

  1. #1
    Junior Member
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    equation set up

    I am having a problem setting up this problem and being successful at solving it.
    A rectangular garden is to be surrounded by a walkway of constant width. The garden's dimensions are 30 ft by 40 ft. The total area, plus garden walkway, is to be 1800ft^2. What must be the width of the walkway to the nearest thousandth.
    I thought A= L X W.
    But I am still stuck and do not get a successful answer.
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, lizard4!

    A rectangular garden is to be surrounded by a walkway of constant width.
    The garden's dimensions are 30 ft by 40 ft.
    The total area, plus garden walkway, is to be 1800 ft².
    What must be the width of the walkway to the nearest thousandth.
    I hope you made a sketch . . .
    Code:
          : x : - -40 - - : x :
          *-------------------* - -
        x |                   |   :
          |   *-----------*   |   :
          |   |           |   |   :
        30|   |30         |   | 30+2x
          |   |    40     |   |   :
          |   *-----------*   |   :
        x |                   |   :
        - *-------------------* - -
          : - - - 40+2x - - - :

    The garden itself is 30 × 40 ft.
    The walkway is x feet wide . . . all the way around.

    The entire region has length (40 + 2x) and width (30 + 2x).

    There is our equation: . (40 + 2x)(30 + 2x) \;=\;1800

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  3. #3
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    A = L x W is formula for area of a rectangle, alright.

    Which rectangle here?

    One rectangle is the 30ft by 40ft garden.
    Another rectangle is that garden that is surrouded by walkways of constant width.

    If c = constant width of the walkways, the other rectangle is
    (30 +2c) by (40 +2c)
    whose area is 1800 sq.ft.

    So,
    (30 +2c)x(40 +2c) = 1800
    30*40 +30*2c +2c*40 +2c*2c = 1800

    ("*" is x or times or multipllied by.)

    1200 +60c +80c +4c^2 = 1800
    4c^2 +140c +1200 -1800 = 0
    4c^2 +140c -600 = 0
    Divide both sides of the equation by 4,
    c^2 +35c -150 = 0

    Cannot factor that. Use the Quadtratic formula,
    c = {-35 +,-sqrt[35^2 -4(1)(-150)]} / 2(1)
    c = {-35 +,-sqrt[1825]} / 2
    c = {-35 +,-42.72001873} /2
    c = -38.860 ft, or 3.860 ft.

    Reject the -38.860 ft because there are no negative dimensions.

    Therefore, the constant width of the walkways is 3.860 ft. ------answer.

    check,
    (3.860)[(30+3.860+3.860)*2 +(40*2)] + 30*40 =? 1800
    600 +1200 =? 1800
    1800 =? 1800
    Yes, so, OK.
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