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Homework Help: Finding the Max Value of a Function on a closed interval

  1. Mar 27, 2008 #1
    [SOLVED] Finding the Max Value of a Function on a closed interval

    1. The problem statement, all variables and given/known data

    Hi, I'm reposting this because it's a subsection to a larger question I had and I figured more people might be able to help with a new topic name.

    anyway i have the equation (eq1) C(t) = C + a(e^.5t) + b(e^.9t) t>0 [3,12]

    a,b and C are constants.


    3. The attempt at a solution

    I realized from long ago that the max and mins of a function occur when y'=0

    So I took the derivative of the function which gave me

    C(t) = Css + a(e^.5t) + b(e^.9t) t>0

    c'(t)= .5a(e^.5t) + .9b(e^.9t) = 0

    i took the ln of the whole thing

    and that gave me

    [ln(.5a)*.5t] + [ln(.9b)*.9t] = 0

    my question now is can use the property that ln(xy) = ln(x)+ln(y) to combine the two ln's and thereby combining the coeffiecients of t such that it is .5t+.9t = .14t

    and the whole thing would be .14t*ln(.45ab)= 0

    ....... that seems to be wrong considering that the t's cancel out but I can't think of much else...
     
  2. jcsd
  3. Mar 27, 2008 #2

    HallsofIvy

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    Not quite. If the max and/or min occur in the interior of an interval, they must occur where y'= 0. The max and/or min may occur at the endpoints of the interval.

     
  4. Mar 27, 2008 #3

    Dick

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    The 'taking the ln of the whole thing' is a completely wrong move. E.g. ln(0)=-infinity and there is no rule that says ln(a+b)=ln(a)+ln(b). Rearrange the equation first. Start by writing .5*a*e^(.5*t)=-.9*b*e^(.9*t). Now move all of the e^ type things over to one side and everything else over to the other. Now 'take the ln of the whole thing'.
     
  5. Mar 27, 2008 #4

    HallsofIvy

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    Not quite. If the max and/or min occur in the interior of an interval, they must occur where y'= 0. The max and/or min may occur at the endpoints of the interval.

    No, no no! Ln(x+ y) is NOT ln(x)+ ln(y)!!
    I recommend factoring out e.5t: e.5t(.5a+ .9be.4t)= 0. Since e.5t is never 0, you must have .5a+ .9be.4t= 0. That should be easy to solve for t. Of course, one of a and b must be negative for this to be true.


    Yes, that is wrong!
     
  6. Mar 27, 2008 #5
    Should've kept those calc books :grumpy: thanks for the help guys
     
  7. Mar 27, 2008 #6
    Just to clarify, I said that the ln(x*y) = ln(x) +ln(y) not ln(x+y)

    that is correct isn't it?
     
  8. Mar 27, 2008 #7

    Dick

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    That's correct. That's what you SAID. If you look back at what you actually USED, it was ln(x+y)=ln(x)+ln(y) (???) when you 'took the ln of the whole thing'.
     
  9. Mar 27, 2008 #8
    yeah, I figured that. It just comes into play right after when I seprate these t's out so i wanted to double check. Thanks again.
     
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