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

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?

• 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.) – littleO Jun 13 '20 at 0:27

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 $$y$$.