# Homework Help: Lagrange again :facepalm:

1. Sep 8, 2010

### naturalnumbas

1. The problem statement, all variables and given/known data
Seems straightforward enough, Lagrangian optimization

2. Relevant equations
Find the max of x^-1 + y^-1 subject to the constraint m=x+y

3. The attempt at a solution

At first I thought no problems, x*=y*=m/2, however:

Using the Lagrangian formula yields derivatives as follows:
wrt x: -x^-2 - lambda
wrt y: -y^-2 - lambda
lambda: m-x-y

Putting the coefficients into a bordered Hessian seems to give a positive def. matrix implying a minimum? Is this a trick question or is it possible to maximize?

2. Sep 8, 2010

### Dick

Yes, x=y=(1/2) is a local minimum. You can also just use your constraint to eliminate y in 1/x+1/y and graph it as a function of x. As x ranges over all possible values you can see there is no global max or global min. There are vertical asymptotes.

3. Sep 8, 2010

### naturalnumbas

Thanks for the reply, you are making sense to me. I'm still confused as to why I was asked to maximize it, when there are no maximum points?

4. Sep 8, 2010

### Dick

Typo? Always a possibility.

5. Sep 8, 2010

### naturalnumbas

I'm thinking you are probably right.

On an exam no less :S.

Guess I should have known better. Thanks again.

6. Sep 9, 2010

### naturalnumbas

Although, I've been reading that extreme values can occur at endpoints, stationary points, and singular points. So obviously this function has some singular points, but I do not know enough to go any further.

Could there be a max at the asymptotes?

7. Sep 9, 2010

### Dick

As I said, graph 1/x+1/(m-x) to see what the function looks like. Yes, it's singular at x=0 and x=m. No maxes there.