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Let's take a simple example $f: \mathbb R^{2} \to \mathbb R$, $f(x,y)=xy$ and then I want to treat $f$ for a constraint $M$ under all possible inequalities:

Case 1) $M:=\{(x,y)\in \mathbb R^{2}|x^2+y^2=1\}$

Case 2) $M:=\{(x,y)\in \mathbb R^{2}|x^2+y^2\leq1\}$

Case 3) $M:=\{(x,y)\in \mathbb R^{2}|x^2+y^2\geq 1\}$

Case 4) $M:=\{(x,y)\in \mathbb R^{2}|x^2+y^2<1\}$

Case 5) $M:=\{(x,y)\in \mathbb R^{2}|x^2+y^2>1\}$

Ideas:

Case 1) This is just the lagrange multiplier method, which is the standard constraint we have used thus far

Case 2) Similar to Case 1) because the Extrema are ALWAYS assumed on the boundaries (question: is this always true?). We simply take the lagrange multipliers of $\partial M$.

Case 3) $M$ is closed, $f$ is continuous and $M$ is bounded below, therefore $f|_{M}$ takes on a minimum (question: is this the correct reasoning?). In order to evaluate the critical values on the open set $M^{\circ}$, I would then simply let $\nabla f(x,y)=\begin{pmatrix} y\\ x \end{pmatrix}=0$, which implies $y=0, x=0$, so $f$ has no critical values on $M^{\circ}$ seeing as though the only critical value $(0,0)\notin M$. Since $f|_{M}$ does take on a minimum, the minimum must therefore be located on the boundary. In order to do this I once again use the lagrange multiplier method. (question: if I had found critical values $(x,y) \in M^{\circ}$, would I still have to compare the lagrange multiplier method used on $\partial M$ with my critical values found on $f|_{M^{\circ}}$ and the largest/smallest values would be global maximum/minimum?)

Case 4) Setting $\nabla f(x,y)=0$, it follows that $y=0, x=0$ and $(0,0) \in M$, therefore we use the hessean matrix to assess whether it is a saddle point, or strict extrema.

Case 5): Since the only critical value $(0,0) \notin M$ and $M$ is open, $f|_{M}$ does not have any extrema.

Answering the questions and identifying any fallacies in my reasoning would be of great help to me!

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Let $M_1,M_2,M_3,M_4,M_5$ be the sets you have defined in cases $1,2,3,4,5$ respectively.

Let's say you have already solved case $1$ using the lagrange multiplier method.

Case 4) $M_4$ is an open set and the function $f(x,y)=xy$ is differentiable in $M_4$, so the only candidates to be extrema are the points $a\in M_4$ such that $Df(a)=0$ (the critical points).

Case 2) "Similar to Case 1) because the Extrema are ALWAYS assumed on the boundaries" this statement is false. However, note that $$M_2=\{(x,y)\in \mathbb R^{2}|x^2+y^2\lt 1\} \cup \{(x,y)\in \mathbb R^{2}|x^2+y^2 = 1\} = M_4 \cup M_{1}$$

You have already found the candidates to be extrema for $M_{1}$ and $M_4$, so the set of candidates to be extrema is just the union of the candidates of $M_1$ and $M_4$.

Case 5) As in case $4$, $M_5$ is an open set and the function $f(x,y)=xy$ is differentiable in $M_5$, so using the same reasoning the candidates to be extrema are the points $a\in M_5$ such that $Df(a)=0$. As you have proved $f|_{M}$ doesn't have any extrema at $M_5$.

Case 3) "$M$ is closed, $f$ is continuous and $M$ is bounded below, therefore $f|_M$ takes on a minimum" I don't really understand what you mean by bounded below. The set $M_3$ is the set of points which are not contained in the unit ball $B((0,0),1)$ so it' s clearly not bounded. However, as in case 2 you have that $M_3 =M_1 \cup M_5$ so the set of possible extrema is the union of the candidates of $M_1$ and $M_5$.

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  • $\begingroup$ Fantastic explanation thank you! One question though regarding: "Similar to Case 1) because the Extrema are ALWAYS assumed on the boundaries". What conditions are needed to be assured that the the extrema lie on the boundary as I have been advised to ignore the interior in some instances but I am not sure when I should do this, and when not. Here for example: math.stackexchange.com/a/2844582/512018 $\endgroup$ – SABOY Jul 11 '18 at 14:18
  • $\begingroup$ I don't know about any condition to assure that other than the function not having critical points at the interior of the set. $\endgroup$ – Yagger Jul 11 '18 at 16:46
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The method of Lagrange multipliers after Joseph-Louis Lagrange was conceived to solve minimization/maximization problems with equality constraints. The case with inequalities can be handled introducing the so called slack variables. In the proposed example, in the inequalities case, the problem can be handled successfully as

$$ x^2+y^2-1 \le 0\to x^2+y^2-1+\epsilon^2 =0\\ x^2+y^2-1 \ge 0\to x^2+y^2-1-\epsilon^2 =0 $$

now focusing on

$$ L = x y + \lambda(x^2+y^2-1+\epsilon^2) $$

The stationary conditions give

$$ \nabla L = \left\{ \begin{array}{rcl} 2 \lambda x+y=0 \\ x+2 \lambda y=0 \\ \epsilon ^2+x^2+y^2-1=0 \\ 2 \epsilon \lambda =0 \\ \end{array} \right. $$

and solving we have

$$ \left[ \begin{array}{ccccc} x & y & \lambda & \epsilon & x y \\ 0 & 0 & 0 & -1 & 0 \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & -\frac{1}{2} & 0 & \frac{1}{2} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{2} & 0 & -\frac{1}{2} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & \frac{1}{2} & 0 & -\frac{1}{2} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & -\frac{1}{2} & 0 & \frac{1}{2} \\ \end{array} \right] $$

Note that the solutions with $\epsilon = 0$ (second to fifth) are those solutions at the feasible region border, while the solutions with $\epsilon \ne 0$ are at the feasible region interior. The qualification of those stationary points should be done analyzing the KKT conditions for each point.

I hope this helps.

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