Maximizing a Function with a Constraint: The Lagrangian Approach

[naturalnumbas]

Homework Statement

Seems straightforward enough, Lagrangian optimization

Homework Equations

Find the max of x^-1 + y^-1 subject to the constraint m=x+y

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?

 

[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.

 

[naturalnumbas]

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.

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?

 

[Dick]

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?

Typo? Always a possibility.

 

[naturalnumbas]

Typo? Always a possibility.

I'm thinking you are probably right.

On an exam no less :S.

Guess I should have known better. Thanks again.

 

[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?

 

[Dick]

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?

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.