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Math Help - I just need these two....

  1. #1
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    May 2005
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    I just need these two....

    4. A new parking lot is twice the area of the old one. If the length of the old one is 100 ft more than the width and the width is the same as it was, what are the dimensions of the lot?

    And now the really hard ones...

    1. The area of a rectangular field is 1200 m^2. Two parallel sides are fenced with aluminum at $20/m. The remaining two sides are fenced with steel at $10/m. If the cost of the fencing is $2200, what is the length of each side fenced with aluminum? What is the length of each side fenced with steel?
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  2. #2
    Junior Member
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    Apr 2005
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    Length of old lot = width + 100
    Area of old lot = width * (width + 100)
    Area of new lot = 2 * width * (width + 100)

    Since the width of both lots are the same, the 2 must go with the length,
    Area of new lot = width *[2(width+100)]
    The length of the new lot is 2*(width+100), the width is the same

    Now the hard one,
    1200 = xy where x is side of aluminum, y is the side with steel

    20(2x) + 10(2y) = 2200 note we multiply by 2 since there are two sides each
    40x + 20y = 2200
    Now, we have a quadratic equation, y = 1200/x
    40x + 20(1200/x) = 2200 multiply everything by x
    40x^2 + 24000 = 2200x
    40x^2 -2200x + 24000 = 0
    Now use the quadratic formula and you get x, then plug in to y = 1200/x
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  3. #3
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    May 2005
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    thank you, thank you, thank you
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