Is Lagrange multipliers and (multivariable) extreme value theorem related?

Question

I couldn't find a question answering this concept but they seem to be related.

Extreme Value Theorem (two variables)

If f is a continuous function defined on a closed and bounded set $A⊂\mathbb{R}^2$, then f attains an absolute maximum and absolute minimum value on A.

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Lagrange Multipliers (two variables)

Extreme values of function f(x, y) subject to constraints g(x, y) = k has solutions in $\nabla f=\lambda \nabla g$.

The constraint in Lagrange Multipliers creates a closed and bounded region that would satisfy EVT, does it not? So does that make Lagrange multipliers a specific case of EVT?

Answer 1

The technique of Lagrange Multipliers cannot be classified as a "special case" of the EVT. Note that the constraint $g$ in an optimization problem may not yield a closed and bounded set to optimize across, so the extreme value theorem may not apply even when the technique of Lagrange multipliers can still find absolute extrema. Take for example the constraint $g(x, y) = x^2 - y^2 = 1$, which is a hyperbola and is certainly not bounded. Nevertheless, absolute extrema may still exist (since the EVT only gives a sufficient condition for there to be an absolute min and max). This is the case when the function we seek to minimize is $$f(x, y) = x^2 + y^2$$ whose absolute minimum on the constraint is attained at $(x, y) = (1, 0)$. You can verify for yourself that there is a solution to $\nabla f = \lambda \nabla g$ with these values of $x$ and $y$.

That being said, in many (maybe even most problems you encounter) both the EVT and Lagrange Multipliers apply.

Answer 2

Well, if you want to relate the theorem to another theorem it is more related to Fermat's theorem. Fermat's theorem states that if a function is differentiable over an interval then the extreme points are either the endpoints or at a point for which the derivative is zero. they are similar because when we want to prove the Lagrange multipliers theorem we parametrize the path and use this theorem. and like how Fermat's theorem can be of help for finding the extrema of the function over an interval the Lagrange's multiplier helps us find the extrema over a constraint region. Hope i helped