1
$\begingroup$

My general question is this:

What are the conditions (if any) such that the method of Lagrange multipliers will NOT find all the critical points of a differentiable function?

To give some context to this very general question, for

$$f(x, y, z) = 600xy + 900xz + 900yz \text{ subject to } xyz = 486$$

I confirmed a minimum at (9, 9, 6) using a Lagrange multiplier. That method also indicated that was the only critical point. However, Wolfram found an approximation to an additional minimum, which looks valid.

So I am confused. My best guess at an explanation is that although the function is everywhere differentiable, the constraint is not continuous everywhere. But that is a pure guess.

To get the full context behind my question, please look at the following thread at a math homework site where I volunteer:

https://www.freemathhelp.com/forum/threads/maximum-minimum-in-multivariable-functions.116663/

$\endgroup$

migrated from mathoverflow.net Jun 17 at 20:27

This question came from our site for professional mathematicians.

  • 2
    $\begingroup$ This would be a great example if it's posted here. $\endgroup$ – Bullet51 Jun 16 at 17:28
  • 7
    $\begingroup$ it seems to be a Mathematica error; if you eliminate $z=486/xy$ the spurious minimum does not appear; probably inquiring at the Mathematica StackExchange site will help. $\endgroup$ – Carlo Beenakker Jun 16 at 17:39
  • 6
    $\begingroup$ It is funny. On the one hand, Wolfram can make quite dazzling computations with special functions and on the other hand it cannot recognize AM-GM when it stares in its face. I guess the hopes/fears that human intelligence will be superseded by AI are highly exaggerated (unless we do a good job unlearning what we know ourselves, that is). Of course, there is only one minimum of $X+Y+Z$ under the conditions $X,Y,Z>0$, $XYZ=1$ (which is equivalent to the initial problem). Why do we start having doubts about something we got with mother's milk when some fancy software outputs something strange? $\endgroup$ – fedja Jun 16 at 18:26
  • 3
    $\begingroup$ @StevenLandsburg Erm... Be careful. This argument is true but it may just mean a third missing minimum. As to the positivity, it is even more obvious than AM-GM that there is no local minimum if two variables are negative (move one negative closer to $0$ and the positive one away from zero keeping their product and the sum will go down). That is just elementary common sense stuff. If one is taught Lagrange multipliers and not explained such things, the whole mathematical education is rather rotten, IMHO. Well, apparently my mood today is not at its best. Apologies if I'm being too blunt. :-) $\endgroup$ – fedja Jun 16 at 20:01
  • 3
    $\begingroup$ "One of the problems with learning a technique is that you rely on the technique rather than intuition." That's exactly my point: when in doubt, stop thinking of fancy things and look at the setup from the perspective of the basic common sense and the techniques you know reasonably well and quite often you'll see the light very quickly :-). $\endgroup$ – fedja Jun 16 at 20:28
1
$\begingroup$

Suppose that $f:\mathbb R^n \to \mathbb R$ and $g:\mathbb R^n \to \mathbb R$ are smooth (to be precise, let's assume they are continuously differentiable), and suppose that $x^\star$ is a local extremum of $f(x)$ subject to the constraint that $g(x) = 0$. If the LICQ constraint qualification $\nabla g(x^\star) \neq 0$ is satisfied (which is usually the case), then we are guaranteed that there exists a Lagrange multiplier $\lambda$ such that $$ \nabla f(x^\star) = \lambda \nabla g(x^\star). $$

In the example problem given in the question, we have $$ g(x_1, x_2,x_3) = x_1 x_2 x_3 - 486. $$ The gradient of $g$ must be nonzero at any point $x$ which satisfies $g(x) = 0$. Thus, any local extremum for the problem given in the question must satisfy the Lagrange multiplier optimality condition. The method of Lagrange multipliers does not fail in this example.

The additional solution found by Wolfram Alpha does not satisfy the Lagrange multiplier optimality condition, so it is not correct. It is not a local extremum.

$\endgroup$
  • $\begingroup$ Thank you. That is a helpful answer, and I have upvoted it. It fully answers my original question: assuming the required conditions are met, every constrained extremum (except presumably at a boundary) will be found by the method of LaGrange multipliers. That leaves two follow-on questions. First, is there a geometric explanation for the first sentence in the next to last paragraph? It makes no intuitive sense to me, but my intuition is admittedly not very reliable outside two dimensions. Because of character limitations, I shall put my second follow-on question in a seperate comment. $\endgroup$ – Jeff Morrow Jun 18 at 16:34
  • $\begingroup$ Second question. I presume that, again given appropriate conditions, the fact that LaGrange multipliers find all constrained extrema does not imply that every point found is indeed an extremum. By analogy to the one independent variable case, extrema have derivatives equal to zero, but derivatives equal to zero can occur at points that are not extrema. Is that correct? We still must consider higher order partials. The point is that I am realizing that I do not understand why LaGrange multipliers work. I know a technique without grasping even an intuitive rationale. $\endgroup$ – Jeff Morrow Jun 18 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.