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Math Help - Area of Rectangle

  1. #1
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    Area of Rectangle

    The initial dimensions of a rectangle are 5x7 cm and the length and width of the reactangle are increasing by the rate of 2 cm/sec. How long will it take for the area to be at least 5 times its initial size?

    I use a calculator to figure out when the size will be equal to or greater to 175 cm squared and the time would be atleast 3 minutes. But how do i figure out the time to the nearest second, the exact time in seconds?
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  2. #2
    No one in Particular VonNemo19's Avatar
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    this help?

    175<(5+2t)(7)

    Where t is time
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  3. #3
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    Quote Originally Posted by chrozer View Post
    The initial dimensions of a rectangle are 5x7 cm and the length and width of the reactangle are increasing by the rate of 2 cm/sec. How long will it take for the area to be at least 5 times its initial size?

    I use a calculator to figure out when the size will be equal to or greater to 175 cm squared and the time would be atleast 3 minutes. But how do i figure out the time to the nearest second, the exact time in seconds?
    InitialArea = 5 x 7 = 35

    Each side increases at the rate of 2cm/sec
    Thus,

    The Area at time zero = (5+2*0)(7+2*0) = 35
    the area at time of 1 second: (5+2*1)(7+2*1) = 7*9 = 63
    the area at time of 2 seconds: (5+2*2)(7+2*2) = 9*11 = 99
    the area at time s will be: (5+2*s)(7+2*s)

    What we seek is a rectangle that is 5 times the initial area.
    5*35 = 175

    the area at time in seconds, s : (5+2*s)(7+2*s) = 175
    expanding:
    5*7 + 10*s + 14*s + 4*s^2 = 175
    simplifying:
    4s^2 + 24s - 140 = 0

    A simple quadratic equation, solve for s.
    The actual answer is between the 3rd second and the 4th second.
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  4. #4
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    Quote Originally Posted by aidan View Post
    InitialArea = 5 x 7 = 35

    Each side increases at the rate of 2cm/sec
    Thus,

    The Area at time zero = (5+2*0)(7+2*0) = 35
    the area at time of 1 second: (5+2*1)(7+2*1) = 7*9 = 63
    the area at time of 2 seconds: (5+2*2)(7+2*2) = 9*11 = 99
    the area at time s will be: (5+2*s)(7+2*s)

    What we seek is a rectangle that is 5 times the initial area.
    5*35 = 175

    the area at time in seconds, s : (5+2*s)(7+2*s) = 175
    expanding:
    5*7 + 10*s + 14*s + 4*s^2 = 175
    simplifying:
    4s^2 + 24s - 140 = 0

    A simple quadratic equation, solve for s.
    The actual answer is between the 3rd second and the 4th second.
    Ok I see... now. Yeah I meant between the 3rd and 4th second not minute.

    Ok I solved it on the calculator and I got around 3.633 seconds. Thanks alot.
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