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Minimization using first differential equation - help

  1. Jan 17, 2008 #1

    suk

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    Hi all
    I am trying to minimize a function by setting the first derivative equal to 0. The strange thing is that I end up with a negative result, which cannot be true (for the application). Any ideas on how this could happen? Do I have an error in my equation somewhere?

    [tex]f(m)=m+n*(\frac{1}{2})^{\frac{m}{n}*ln(2)}[/tex]

    (Proceeding with Differentiation on m)[tex]\Rightarrow[/tex]

    [tex]\frac{df(m)}{dm}=1+n*(\frac{1}{2})^{\frac{m}{n}*ln(2)}*ln(\frac{1}{2})*\frac{ln(2)}{n}[/tex]

    [tex]\frac{df(m)}{dm}=1+(\frac{1}{2})^{\frac{m}{n}*ln(2)}*ln(\frac{1}{2})*ln(2)[/tex]

    (Equal to zero for Minimization) [tex]\Rightarrow[/tex]

    [tex]0=1+(\frac{1}{2})^{\frac{m}{n}*ln(2)}*ln(\frac{1}{2})*ln(2)[/tex]

    [tex]1=-(\frac{1}{2})^{\frac{m}{n}*ln(2)}*(-ln(2))*ln(2)[/tex]

    [tex]1=(\frac{1}{2})^{\frac{m}{n}*ln(2)}*ln(2)*ln(2)[/tex]

    [tex]1=(\frac{1}{2})^{\frac{m}{n}*ln(2)}*(ln(2))^{2}[/tex]

    [tex](\frac{1}{(ln(2))^{2}})=(\frac{1}{2})^{\frac{m}{n}*ln(2)}[/tex]

    (Substituting values for [tex](ln(2))^{2})\Rightarrow[/tex]

    [tex](\frac{1}{0.480453015})=(\frac{1}{2})^{\frac{m}{n}*ln(2)}[/tex]

    [tex]2.08136898=(\frac{1}{2})^{\frac{m}{n}*ln(2)}[/tex]

    (Taking natural logarithm on both sides)[tex] \Rightarrow[/tex]

    [tex]ln(2.08136898)=(\frac{m}{n}*ln(2))*(ln(\frac{1}{2})) [/tex]

    (Substituting values)[tex]\Rightarrow[/tex]

    [tex]0.733025841=(\frac{m}{n}*0.693147181)(-0.693147181)[/tex]

    [tex](\frac{m}{n})=-1.52569724[/tex]

    The result I need is the ratio between m and n (the LHS in the above equation). But, the practical application of this equation doesn't make sense if the ratio is negative. Am I missing something?
     
  2. jcsd
  3. Jan 17, 2008 #2

    nicksauce

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    This is not a physical result, so it seems there are no maxima or minima in the domain of your function m,n > 0. Therefore you should check the limits m->0 and m->infinity for the maximum and minimum.
     
  4. Jan 17, 2008 #3

    suk

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    Minimizing subject to inequality constraints

    Hi all.
    I need some help on how to minimize the following equation subject to inequality constraints:
    The final result is I need a relation between m and n that minimizes the equation and satisfies the constraints.
    Given equation:
    [tex] f(m,n)=m+((\frac{1}{2})^{\frac{m}{n}*ln(2)}*n)[/tex]
    Constraints:
    m>=0
    n>=0

    Any suggestions?
     
  5. Jan 18, 2008 #4

    suk

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    But the limits for the above equation at infinity become undefined. Or am I wrong? How do I then get to the minima?
     
  6. Jan 18, 2008 #5

    arildno

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    There are no local extrema within the open domain, since you have found that that would imply m/n<0, i.e, any local extremum lies outside the required domain.

    Therefore, an extremal value within the region can only lie on the border of the region.

    One border is n=0.

    At that border, we have f(m,0)=m, having a minimum at m=0, i.e, (0,0) is an extremum at the border. The other border gives the same extremum.

    Thus, (0,0) is the only extremum for f, yielding 0 as its minimum value.
     
  7. Jan 18, 2008 #6

    HallsofIvy

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    They can "become undefined" by going to +infinity or -infinity. If they go to + infinity, there cannot be a maximum. If they go to -infinity, there cannot be a minimum. If they do not, if they are bounded, there will "least upper bound" and "greatest lower bound" but not necessarily a maximum or minimum.
     
  8. Jan 18, 2008 #7

    HallsofIvy

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    This same question was also posted under "differential equations". Since it has nothing to do with "differential equations" (you are using the "first derivative", not "first differential equation"), I am merging the two threads here.
     
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