0
$\begingroup$

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.

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?

Improve this question
$\endgroup$
  • 1
    $\begingroup$ The Lagrange multiplier optimality condition is analogous to the optimality condition $\nabla f(x) = 0$ for unconstrained optimization. In both cases, for a minimization problem, the optimality condition can be interpreted as saying that if you move a short distance in a feasible direction, the objective function value does not decrease. (It's not totally obvious at first that the Lagrange multiplier optimality condition can be interpreted in this way. The book Numerical Optimization by Wright et al has a good explanation. I've also explained this in other questions if you look at my history.) $\endgroup$ – littleO Jun 13 '20 at 0:27
1
$\begingroup$

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.

Improve this answer
$\endgroup$
  • $\begingroup$ makes sense: all the examples I've seen have indeed been circles. I get it now, thanks! $\endgroup$ – user29418 Jun 13 '20 at 22:14
1
$\begingroup$

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

Improve this answer
$\endgroup$

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.