Lagrange multiplier conundrum

[hellosoupy]

Hi! I went to my professor with this problem, and he found a work-around, but he was kind of stumped too. Here's the problem:

Find the minimum value of f(x,y)=x^2-y^2 subject to the constraint x^2+y^2=4

This is in a chapter that has as its subtitle: "the method of lagrange multipliers". And yet, from the first order condition of x, you get a different value of lambda (lambda = 1) than from the first order condition of y (lambda = -1). Does this mean that you can't use the method of lagrange multipliers for this problem, and if you can't, why not? If you can use the method of lagrange multipliers, how do you reconcile these 2 different values of lambda?

Thanks in advance!

 

[Steven G]

We do not solve problems for students on this forum. We prefer to see the student's work and then discuss what is going on. Did you forget to read the forum's posting guidelines??
Please post back showing the work which you have done so we can see what is going on.

 

[JeffM]

[MATH]L(x,\ y,\ \lambda ) = f(x) - \lambda(4 - x^2 - y^2) = x^2 - y^2 - \lambda(4 - x^2 - y^2) \implies[/MATH]
[MATH]\dfrac{\delta L}{\delta x} = 0 \implies 2x - 2\lambda x = 0 \implies x(1 - \lambda) = 0.[/MATH]
[MATH]\dfrac{\delta L}{\delta y} = 0 \implies -2y - 2\lambda y = 0 \implies y(1 + \lambda) = 0.[/MATH]
[MATH]\dfrac{\delta L}{\delta \lambda} = 0 \implies 4 - x^2 - y^2 = 0 \implies x^2 + y^2 = 4.[/MATH]
Now what you say is NOT QUITE RIGHT. Your professor was having a brain freeze (it happens to us all). You can use the method of LaGrangian multipliers. There is no inconsistency.

[MATH]2x(1 - \lambda) = 0 \implies x = 0 \text { OR } \lambda = 1.[/MATH]
[MATH]2y(1 + \lambda) = 0 \implies y = 0 \text { OR } \lambda = -1.[/MATH]
So lambda must have one of two different values. We are dealing with a quadratic. Why does having two possible solutions feel familiar? There would be a problem of consistency only if both x and y were not equal to 0.

But that is impossible because [MATH]x^2 + y^2 = 4 \ne 0^2 + 0^2.[/MATH]
You have to look at the whole system.

So we have four solutions (not surprising when we have two squared terms).

[MATH]x = 0 \implies 0^2 + y^2 = 4 \implies y = 2 \text { or } - 2.[/MATH]
[MATH]x \ne 0 \implies \lambda = 1 \implies (1 + \lambda) = 2 \implies 2y = 0 \implies[/MATH]
[MATH]y = 0 \implies x^2 = 4 \implies x = 2 \text { or } x = - 2.[/MATH]
But it is obvious that if x is not zero and y = 0

[MATH]f(x,\ y) = (\pm2)^2 - 0^2 = 4.[/MATH]
If y is not zero and x = 0

[MATH]f(x,\ y) = 0^ 2 - (\pm2)^2 = - 4.[/MATH]
So where is f(x, y) minimum while satisfying the constraint?

 

[hellosoupy]

Thank you Jeff. That's brilliant. f(x,y) is at a minimum at (0,2) and (0,-2).

Nowhere in the chapter did it talk about the possibility of lamba having 2 different values (in the same problem), so it was surprising when 2 of the earliest problems in the book had this outcome. My professor got this answer by subbing the constraint into the function and by using intuition. Your method feels more comprehensive, though.

Thanks so much for your help!

 

[JeffM]

Yes, you got it.

I thank you for the compliment, but it is completely undeserved: it was LaGrange who was brilliant. There is a street named for him in Turin. I once spent a morning at a coffee shop on that street waiting for my wife.

A few follow-up points.

I do not remember seeing a formal proof, but I believe that any problem solvable by LaGrangian multipliers can be solved by your professor’s method. It is just easier to use multipliers, particularly if there are multiple constraints. Your professor’s method, however, may be more general.

Whenever there are multiple extrema subject to a constraint, there may be multiple values of the variable associated with that constraint. (I do not know whether or not there MUST be multiple values.)

 

[JeffM]

I realize there was an error in my first post. It was a harmless error. But still.

We say that all three differentials will simultaneously = 0 if and only if

[MATH]\{x = 0 \lor \lambda = 1\} \land \{y = 0 \lor \lambda = - 1\} \land \{x^2 + y^2 = 4\}.[/MATH]
Now this would be an impossible condition if it said

[MATH]\{\lambda = 1\} \land \{\lambda = - 1\}.[/MATH]
Which is what stopped you initially. But that is not the condition.

The first two parts of the condition can be met if

[MATH]x = 0, \ y = 0,[/MATH]
[MATH]x = 0, \ y \ne 0, \ \lambda = - 1,[/MATH] or

[MATH]x \ne 0,\ y = 0, \lambda = 1[/MATH].

The first of those alternatives, however, is inconsistent with [MATH]x^2 + y^2 = 4.[/MATH]
The possibility of [MATH]x \ne 0 \land y \ne 0[/MATH]
entails [MATH]1 = \lambda = - 1 \implies 1 = - 1, \text { which is absurd.}[/MATH]
We can logically exclude the possibility that seemingly creates an inconsistency.

I did not clearly explain this in my first answer.

 

[hellosoupy]

Thanks so much for your follow-up comments. I'm struggling with the meaning of the following:
"Whenever there are multiple extrema subject to a constraint, there may be multiple values of the variable associated with that constraint. (I do not know whether or not there MUST be multiple values.)"

I've thought of it as the function being subject to the constraint and am having trouble conceptualizing extrema being subject to a constraint. I think of the constraint as a box within a space (the space constituting the function) that defines and/or contains the extrema. Is that wrong?

I'm also having trouble conceptualizing the second part: "there may be multiple values of the variable associated with that constraint". Are you saying here that there may be multiple values for lambda? or for x and/or y also?

Regarding your second follow-up post: So well-put. Thanks!

 

[JeffM]

You are taking me into deep waters. Any helper who wants to contribute or correct any error I may make should feel free to do so.

If we are being exact, what is always being constrained in the first instance is the domain, and that may or may not affect the placement and value of one or more local extrema. The function is not altered in any way except as to domain. The locations and hence the values of local extrema may or not be affected by the limitation of the domain. If the local extrema are not affected in any way, then it is somewhat misleading to say that they are constrained. If, on the other hand, any local extrema are created or altered by the constraint on the domain, it may be a little loose to say that the extrema are constrained, but it is not misleading.

 

[hellosoupy]

Okay. I think I've got that part. Thanks!

 

[JeffM]

Your other question has me somewhat perplexed for an answer.

The multipliers are variables rather than numbers and so can take on different values. Moreover, they are meaningful only close to boundaries.

You might want to think about what it means when a LaGrangian multiplier equals zero.

 

[hellosoupy]

That makes total sense now. I wasn't thinking of lambda as a variable but that's totally what it is. Thanks!