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so this is an exmaple from a book I have.

Find Max/Min of $f=\frac{x^2}{2} + \frac{y^2}{2}$ on $E=\big\{ (x,y)\in \mathbb R \big| \frac{x^2}{2} + y^2 \leq 1\big\}$

1. Step: We look at the Interior of E:

$df=(x,y)=0 \Rightarrow p_0 = (0,0)$

2. Step: We look at $\partial E$

$g(x,y):=\frac{x^2}{2}+y^2 - 1 = 0 $

We define the lagrangian function:

$L:=f-\lambda g = \frac{x^2}{2}+\frac{y^2}{2}-\lambda (\frac{x^2}{2}+y^2-1)$

We solve

$\frac{\partial L}{\partial x}=x -\lambda x = 0 \quad \Rightarrow \quad x=0, \lambda = 1$

$\frac{\partial L}{\partial y}=y-2y=0\quad \Rightarrow \quad y=0, \lambda = -\frac{1}{2}$

With $g$ we get the candidates: $p_{1,2}=(0,\pm 1)$ and $p_{3,4}=(\pm\sqrt{2},0)$

Step 3: We evaulate $f(p_i)$

$f(0,0)=0$

$f(\pm\sqrt{2},0)=1$

$f(0,\pm 1) = \frac{1}{2}$

So the global min is $(0,0)$ and the global max are $(\pm\sqrt{2},0)$

Question 1: What can we say about the points $(0,\pm 1)$?

Question 2: If we wanted to look at all points and at local ones as well we would need hessian matrix, no?

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Stating the problem as

$$ L(x,y,\lambda,\epsilon) = \lambda \left(\epsilon ^2+ \frac{x^2}{2}+y^2-1\right)+\frac{1}{2} \left(x^2+y^2\right) $$

we have the stationary condition

$$ \nabla L = (L_x,L_y,L_{\lambda},L_{\epsilon}) = 0 $$

or

$$ \lambda x+x = 0\\ 2 \lambda y+y = 0\\ \epsilon ^2+\frac{x^2}{2}+y^2-1 = 0\\ \epsilon \lambda = 0 $$

Solving we get the set

$$ x = 0, y = \pm 1, \lambda = -1/2, \epsilon = 0\\ x = \pm\sqrt2, y = 0, \lambda = -1, \epsilon = 0 $$

those points are located at the boundary ($\epsilon = 0$) and the second of them describes the maximum solutions. See also $\lambda < 0$

and

$$ x = 0, y= 0, \lambda = 0, \epsilon = 1 $$

This point is the minimum point and is internal to the boundary region $\epsilon \ne 0$

Attached a plot showing the maximum set (red) and the minimum point (blue).

enter image description here

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  • $\begingroup$ What's the idea of the epsilon? $\endgroup$ – xotix May 5 '18 at 12:47
  • $\begingroup$ The $\epsilon$ is a slack variable to transform the inequality restriction into an equality $\frac{x^2}{2}+y^2 \le 1 \equiv \frac{x^2}{2}+y^2 - 1+\epsilon^2 = 0$ $\endgroup$ – Cesareo May 5 '18 at 13:09

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