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Homework Help: Exponential growth problem (at least I think it is)

  1. May 27, 2009 #1
    1. The problem statement, all variables and given/known data

    The measure of effectiveness of a sleep deprived student studying for a test on a scale of 0 to 1 is given by the formula M(t) = ln(t^2) / 3t, where 't' is the time in hours that a student spends studying, 0 < t ≤ 4.

    Determine the maximum measure of effectiveness of studying and the time at which it is reached.

    3. The attempt at a solution

    In all honest I've got no idea what going on here, this was just an easter-egg question our teacher handed out at the end of class for a few extra marks. We haven't covered this in class yet, and I'm trying to absorb as much information as possible from my utterly useless textbook, but it's not going very well. It's all just a jumble of balderdash to me at the moment! lol!

    If anyone could help me figure this out I'd really appreciate it! I gotta keep my average at 91% of else my university acceptance gets withdrawn! every mark helps!

    Thanks again everyone!
  2. jcsd
  3. May 27, 2009 #2


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    The idea is to maximise M with respect to t.

    To do this, you can either plot a graph of M against t, or use differentiation.
  4. May 27, 2009 #3
    Is the derivative

    M'(t) = 6 - 3ln(t^2) / 9t^2

  5. May 27, 2009 #4
    No. Use the quotient rule.
  6. May 27, 2009 #5
    I did, and that's what I got. Did I miss something somewhere?
  7. May 27, 2009 #6
    No, I'm wrong. Sorry. I thought your answer was [tex] 6 - \frac {3ln(t^{2})}{9t^{2}}[/tex]. :redface:
  8. May 27, 2009 #7


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    Homework Helper

    All you missed were some parentheses:
    [tex]M'(t) = (6 - 3\ln(t^2))\ /\ (9t^2)[/tex]
    I checked with Mathematica to make sure it's right.
  9. May 27, 2009 #8
    So now that I've got my trusty derivative. Where do I go from here?
  10. May 27, 2009 #9
    Set it equal to zero and solve for t. You''ll get two answers, but one isn't in your domain.
  11. May 27, 2009 #10


    Staff: Mentor

    From a graph of this function, I see a maximum somewhere between 2.6 and 2.75. BTW, this is not exponential growth, nor even logarithmic growth. I don't think there's a name for this kind of function.
  12. May 27, 2009 #11
    Well I ended up getting t = 2.72, and subbing that into the original equation I get M(t) = 0.245.

    So hopefully its correct, and I'll post back and let ya'll know how it went! Thanks again everyone for the help!
  13. May 27, 2009 #12
    You don't have to round. The answer is e.
  14. May 27, 2009 #13
    [tex] \frac {6 - 3ln(t^{2})}{9t^{2}} = 0 [/tex]

    [tex] 6- 3ln(t^{2}) = 0 [/tex]

    [tex] ln(t^{2}) = 2 [/tex]

    [tex] t^{2} = e^{2} [/tex]

    [tex] t = e [/tex] (since -e is not between 0 and 4)
    Last edited: May 27, 2009
  15. May 27, 2009 #14
    You can use the second derivative test to check that e is indeed a maximum. Find the second derivative of M(t) and evaluate it at e. If it's less than zero, then e is a maximum.
    Last edited: May 27, 2009
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