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Math Help - Quick question about Optimization

  1. #1
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    Quick question about Optimization

    "A rumour spreads through a population in such a way that t hours after its beginning, the percentage of people involved in passing the rumour is given by P(t) = 100(e^(-1) - e^(-4t)). What is the highest percentage of people involved in spreading the rumour within the first 3 h? When does this occur?"

    Given the equation, P(t) = 100(e^(-1) - e^(-4t)), this is what I've done so far:

    P(t) = 100(e^(-1) - e^(-4t))
    P(t) = (100/e) - (100e^(-4t))
    P'(t) = -100e^(-4t)(-4)
    P'(t) = 400e^(-4t)

    Where do I go from this point? If I make the equation equal to zero, I won't be able to solve for the time.

    Thanks.


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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jeavus View Post
    "A rumour spreads through a population in such a way that t hours after its beginning, the percentage of people involved in passing the rumour is given by P(t) = 100(e^(-1) - e^(-4t)). What is the highest percentage of people involved in spreading the rumour within the first 3 h? When does this occur?"

    Given the equation, P(t) = 100(e^(-1) - e^(-4t)), this is what I've done so far:

    P(t) = 100(e^(-1) - e^(-4t))
    P(t) = (100/e) - (100e^(-4t))
    P'(t) = -100e^(-4t)(-4)
    P'(t) = 400e^(-4t)

    Where do I go from this point? If I make the equation equal to zero, I won't be able to solve for the time.

    Thanks.


    first off. your derivative is wrong. you have to distribute the 100 to both terms. secondly, you are looking for the absolute max i suppose. if there is no extrema to test, check the end points. this function is clearly increasing, so it should be at t = 3
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    first off. your derivative is wrong. you have to distribute the 100 to both terms. secondly, you are looking for the absolute max i suppose. if there is no extrema to test, check the end points. this function is clearly increasing, so it should be at t = 3

    I did distribute 100 to both terms.

    The book gives these answers: 47.25% when t = 0.46h.

    If that helps at all.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jeavus View Post

    I did distribute 100 to both terms.

    The book gives these answers: 47.25% when t = 0.46h.

    If that helps at all.
    How sly you are, Jeavus, you changed it just before i clicked the "quote" button. I know what i saw, i am not crazy

    anyway, that cannot be right. The graph does not even reach 40 in the first 3 hours. see the graph below: the red line is the graph, the green line is where t = 3. Note that the graph does not pass 40

    check to see that you asked the right question
    Attached Thumbnails Attached Thumbnails Quick question about Optimization-graph.jpg  
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  5. #5
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    Haha, you're correct in the edit. :P

    Anyway, I am checking the right question, I guess the book is wrong for this one.
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