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I am reviewing some basic calculus and am puzzled by introductory texts' justification for the Lagrange Multiplier Method, which is usually an argument about parallel gradients at extrema. But consider this example:

Let $f(x,y) = y^2$ and $g(x,y) = x - y$. Minimize $f$ with respect to the constraint $g(x,y) = 0$.

Now clearly $f(0,0) = 0$ is minimum, but at this point $\nabla f = (0,0), $ whereas $\nabla g = (1, -1)$ and so the justification of the Lagrange Multiplier Method--that the tangent planes are parallel at extrema--fails to hold true here.

I am certainly missing some simple condition required to apply the method. Could anyone set me straight or point me towards a (hopefully self-contained) reference? Thanks!

Note: the above is similar to this question but its accepted answer doesn't address the parallel-ness of gradients.

JosephSlote
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    The method fails at points where either gradient vanishes. You have to examine such points separately—they may or may not correspond to constrained extrema. – amd Sep 10 '19 at 19:00
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    Because the gradient of $f$ vanishes at a point, it is a non-constrained critical point. Such a point certainly counts as a constrained critical point, as well. If $\nabla g$ vanishes, then we have a more serious issue, because the constraint set is likely to be non-smooth there. – Ted Shifrin Sep 10 '19 at 19:04

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