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Math Help - D(t) = (4t)/ (0.01t^2 +5.1)

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    D(t) = (4t)/ (0.01t^2 +5.1)

    For these I'm clueless. Please help

    5) Function D(t) = (4t)/(0.01t^2+5.1). Given the concentration, D, of a drug in bloodstream where D is measured in micrograms per milliliter and t is minutes from the time the drug is taken. What's the maximum concentration of the drug found in bloodstream at any time?
    A) 0.01 micrograms/ milliliters
    B) 4 micrograms/ milliliters
    C) 5.1 micrograms/ milliliters
    D) 8.856 micrograms/ milliliters
    E) 22.583 micrograms/ milliliters
    Last edited by mr fantastic; April 23rd 2011 at 12:48 AM. Reason: Deleted excess questions, re-titled.
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by pcalhelp View Post

    5) Function D (t) = (4t)/ (0.01t square+5.1). Given the concentration, D, of a drug in bloodstream where D is measured in micrograms per milliliter and t is minutes from the time the drug is taken. What's the maximum concentration of the drug found in bloodstream at any time?
    A) 0.01 micrograms/ milliliters
    B) 4 micrograms/ milliliters
    C) 5.1 micrograms/ milliliters
    D) 8.856 micrograms/ milliliters
    E) 22.583 micrograms/ milliliters
    for this you can do the following:
    D(t) = (4t)/[(0.01t)^2 +5.1] = (4)/[0.01t + (5.1/t)]
    to find maximum of (4)/[0.01t + (5.1/t)] you need to find the minimum of 0.01t + (5.1/t). Use AM-GM inequality.
    did this help?
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    Quote Originally Posted by abhishekkgp View Post
    for this you can do the following:
    D(t) = (4t)/[(0.01t)^2 +5.1] = (4)/[0.01t + (5.1/t)]
    to find maximum of (4)/[0.01t + (5.1/t)] you need to find the minimum of 0.01t + (5.1/t). Use AM-GM inequality.
    did this help?
    How do you find the minimum? What is AM-GM?
    Sorry, I;m not very good at math
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  4. #4
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by pcalhelp View Post
    How do you find the minimum? What is AM-GM?
    Sorry, I;m not very good at math
    AM-GM inequality states that if a>0 and b>0 then a+b >= 2(sqrt(ab))


    you could also check this out Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia
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    "AM-GM": Arithmetic Mean- Geometric Mean. The arithmetic mean of two numbers is (a+ b)/2. The geometric mean of two numbers is \sqrt{ab}. The "Arithmetic Mean-Geometric Mean inequality" says that the arithmetic mean is always less than equal to the geometric mean.

    (More geherally, the arithmetic mean of n numbers, a_1, a_2, \cdot\cdot\cdot, a_n is \frac{a_1+ a_2+ \cdot\cdot\cdot+ a_n}{n} and the geometric mean is \sqrt[n]{a_1a_2\cdot\cdot\cdot a_n}. There is also the "harmonic mean". The harmonic mean of 2 numbers is \frac{2}{\frac{1}{a}+ \frac{1}{b}}. The harmonic mean of n numbers is \frac{n}{\frac{1}{a_1}+ \frac{1}{a_2}+ \cdot\cdot\cdot+ \fra{1}{a_n}}
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    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    The "Arithmetic Mean-Geometric Mean inequality" says that the arithmetic mean is always less than equal to the geometric mean.
    hey halls!
    there's a typo error in there!
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  7. #7
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    Quote Originally Posted by abhishekkgp View Post
    hey halls!
    there's a typo error in there!
    What, only one???
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  8. #8
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    What, only one???
    i can see only one. which is(are) the other one(ones)??
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