Constrained Extrema: How to find end points of multivariable functions for global extrema

Question

Example: Let $f:\mathbb R^{2}\to\mathbb R$, $f(x,y)=xy$ and $h(x,y)=x^2+y^2-1$

Task: Find the local extrema AND global extrema of under the constraint of $h=0$.

For $g_{\lambda}:=f-\lambda h$, I found a case $\lambda=\frac{1}{2}$ and another $\lambda=\frac{-1}{2}$

Case 1: $\lambda = \frac{1}{2}$

It follows that we have possible extrema $(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}),(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$

Case 2: $\lambda=-\frac{1}{2}$

It follows we have possible extrema $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}),(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$

Since $\{(x,y)\in \mathbb R^{2}:x^2+y^2=1\}$ is a compact set, $f$ is bounded under the constraint $h=0$ and it takes on a global maximum and minimum.

Question: Through Case 1 & Case 2, we found four critical values §, but from what I remember from single variable calculus, we also need to evaluate $f$ at its end points under the constraint $h=0$, and compare these endpoints with §.

How exactly would I find these endpoints in $f$ when it is multivariable

Answer 1

There are no endpoints here. That's for the finding global extrema of functions defined on intervals of the type $[a,b]$.

Answer 2

Hint: You can reduce your Problem to one variable: $$h(x,\pm\sqrt{1-x^2})=x(\pm\sqrt{1-x^2})$$

Answer 3

With constrain $x^2+y^2=1$ we see that the absolute minimum is $-\dfrac12$: $$f=\dfrac12\left(2xy+0\right)=\dfrac12\left(2xy+x^2+y^2-1\right)=\dfrac12(x+y)^2-\dfrac12\geqslant-\dfrac12$$