# Minimization using first differential equation - help

1. Jan 17, 2008

### 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?

2. Jan 17, 2008

### 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.

3. Jan 17, 2008

### 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?

4. Jan 18, 2008

### suk

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

5. Jan 18, 2008

### 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.

6. Jan 18, 2008

### HallsofIvy

Staff Emeritus
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.

7. Jan 18, 2008

### HallsofIvy

Staff Emeritus
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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