Limitations on method of Lagrange multipliers 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/ 
 A: 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. 
