Minimization using first differential equation - help

[suk]

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?

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

(Proceeding with Differentiation on m)\Rightarrow

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

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

(Equal to zero for Minimization) \Rightarrow

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

1=-(\frac{1}{2})^{\frac{m}{n}*ln(2)}*(-ln(2))*ln(2)

1=(\frac{1}{2})^{\frac{m}{n}*ln(2)}*ln(2)*ln(2)

1=(\frac{1}{2})^{\frac{m}{n}*ln(2)}*(ln(2))^{2}

(\frac{1}{(ln(2))^{2}})=(\frac{1}{2})^{\frac{m}{n}*ln(2)}

(Substituting values for (ln(2))^{2})\Rightarrow

(\frac{1}{0.480453015})=(\frac{1}{2})^{\frac{m}{n}*ln(2)}

2.08136898=(\frac{1}{2})^{\frac{m}{n}*ln(2)}

(Taking natural logarithm on both sides)\Rightarrow

ln(2.08136898)=(\frac{m}{n}*ln(2))*(ln(\frac{1}{2}))

(Substituting values)\Rightarrow

0.733025841=(\frac{m}{n}*0.693147181)(-0.693147181)

(\frac{m}{n})=-1.52569724

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?

 

[nicksauce]

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.

 

[suk]

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:
f(m,n)=m+((\frac{1}{2})^{\frac{m}{n}*ln(2)}*n)
Constraints:
m>=0
n>=0

Any suggestions?

 

[suk]

But the limits for the above equation at infinity become undefined. Or am I wrong? How do I then get to the minima?

 

[arildno]

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.

 

[HallsofIvy]

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.

 

[HallsofIvy]

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.