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Math Help - Volume

  1. #1
    Junior Member Dragon's Avatar
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    Volume

    New Age special Effects Inc prepares computer software based on specifications prepared by film directors. To simulate an approaching vehicle they begin with a computer image of a 5-cm by 7-cm by 3-cm box. The program increases each dimension at a rate of 2cm/sec.How long does it take for the volume V of the box to be at least 5 times its initial size?
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  2. #2
    MHF Contributor
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    What an odd question. "At least"? There are infinitely many answers.

    "t" is the number of seconds to reach EXACTLY five times,

    (5+2t)(7+2t)(3+2t) = 5*(5*7*3)

    Good luck. I suspect you'll need numerical methods.
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  3. #3
    Junior Member Dragon's Avatar
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    How do u solve that?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TKHunny View Post
    What an odd question. "At least"? There are infinitely many answers.

    "t" is the number of seconds to reach EXACTLY five times,

    (5+2t)(7+2t)(3+2t) = 5*(5*7*3)

    Good luck. I suspect you'll need numerical methods.
    Quote Originally Posted by Dragon View Post
    How do u solve that?
    Yep, it's an ugly one.

    The simplest way is (if you have a graphing calculator) to graph it and use the graphing window to estimate the root for you.

    If you have Calculus you could try the Newton-Raphson method.

    If you are sado-masochistic you could try Cardano's Method to find the exact root.

    -Dan
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  5. #5
    MHF Contributor
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    Smile

    "How long does it take for the volume V of the box to be at least 5 times its initial size?"

    If you want to answer just that, then

    Volume at t=0, V(0) = 5*7*3 = 105 cc
    5 times that is 5(105) = 525 cc

    V at t=1, V(1) = (5+2)(7+2)(3+2) = 7*9*5 = 315 cc
    V at t=2, V(2) = (7+2)(9+2)(5+2) = 9*11*7 = 693 cc ---greater than 525 cc already.
    So, 2 seconds ---------------answer.

    If you want to get the exact time when V(t) = 525 cc, do some iteration.

    So t must be between 1 second and 2 seconds.

    Try t = 1.5 sec, or 0.5 sec from 1 sec,
    The changes in each dimensions, 2 cm/sec, must be linear only, so in 0.5 second, the increase is 1 cm only. Thus,
    V(1.5) = (7+1)(9+1)(5+1) = 8*10*6 = 480 cc ----still less than 525 cc.
    So t is between 1.5 sec and 2 sec.

    Try t = 1.75 sec. or 0.75 sec after 1 sec,
    The changes then in each dimension is (2cm /1sec)(0.75sec) = 1.5 cm. Thus,
    V(1.75) = (7+1.5)(9+1.5)(5+1.5) = (8.5)(10.5)(6.5) = 580.125 cc, which is already greater than 525 cc.
    So t is between 1.5 sec and 1.75 sec.

    Try t = 1.6seconds......etc.
    V(1.6) = (7+1.2)(9+1.2)(5+1.2) = 518.568 -----still less than 525
    So a bit higher t

    Try t = 1.62 sec
    V(1.62) = (7+1.24)(9+1.24)(5+1.24) = 526.516 ----almost.

    Try t = 1.617 sec
    V(1.617) = (7+1.234)(9+1.234)(5+1.234) = 525.319 ----

    -------------------
    If you are used to that kind of iteration, it is fast.
    No need to know Calculus in that iteration.
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