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Consider the constraint $$S_1 = \{(x, y) \; |\; \sqrt{x} + \sqrt{y} = 1 \}$$ How to use Lagrange Multipliers, when the constraint surface has a boundary?

In this case, after the Lagrange multiplier method gives candidates for maxima/minima, we need to check the "boundary points" of $S_1$, namely, $(1,0)$ and $(0,1)$ to get the global max/min. I can see that these two are "boundary points" intuitively when I plot the curve.

However, instead if the constraint set be
$$S_2 = \{ (x, y) \; |\; x^2 + y^2 = 1\},$$ then in this question, one answer states that for this constraint set, there is no "boundary point". Constrained Extrema: How to find end points of multivariable functions for global extrema

The only difference I see is that pictorially, one is a closed curve, but the other is not.

However, I am unable to see what is the mathematical definition that will allow me to conclude that $S_1$ has boundary points $(0, 1)$ and $(1,0)$ and $S_2$ has none?

Q) What is the definition of "end point" or "boundary point" being used here that explains both $S_1$, $S_2$.

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In many extremal problems the set $S\subset{\mathbb R}^n$ on which the extrema of some function $f$ are sought is stratified, i.e., consists of points of different nature: interior points, surface points, edges, vertices. If an extremum is assumed in an interior point it comes to the fore as solution of the equation $\nabla f(x)=0$. An extremum which is at a (relative) interior point of a surface or an edge comes to the fore by Lagrange's method or via a parametrization of this surface or edge. Here (relative) interior refers to the following: Lagrange's method deals only with constrained points from which you can march in all tangent directions of the submanifold (surface, edge, $\ldots$) defined by the constraint(s), all the while remaining in $S$. Now at a vertex there are forbidden marching directions on all surfaces meeting at that vertex. If the extremum is taken on such a vertex it only comes to the fore if you have deliberately taken all vertices into your candidate list.

Now your $S_1$ is an arc in the plane with two endpoints. (The latter are not immediately visible in your presentation of $S_1$, but you have found them.) Your candidate list then should contain all relative interior points of the arc delivered by Lagrange's method plus the two boundary points.

The circle $S_2\!: \ x^2+y^2=1$ however has only "interior" points. The candidate list then contains only the points found by Lagrange's method.

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  • $\begingroup$ Does this mean that if the level set is not a closed curve, then I will have a boundary point ? $\endgroup$
    – me10240
    May 27 '19 at 8:46
  • $\begingroup$ The level set could also extend to infinity. It means that you should have a qualitative overview over the set $S$ and its kinds of points. $\endgroup$ May 27 '19 at 8:57
  • $\begingroup$ Is there a mathematical way to describe what you said-- March in all tangent directions .... I am confused whether it refers to the curve ( which after parametrization is 1d, or the multivariable function representing it $\endgroup$
    – me10240
    May 27 '19 at 9:06
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If the constraint set is defined as the set of points where $g(x,y)=0$,then its 'boundary points' will be those points where $\frac{\partial g}{\partial x}$ or $\frac{\partial g}{\partial y}$ is undefined.

Lets suppose that the constraint set is $\{x,y||x|+|y|=1\}$, so we want so maximise $f(x,y)$ subject to the constraint $g(x,y)=|x|+|y|-1=0$.

We do this by defining the Lagrangian $\mathcal{L}=f-\lambda g$ and examining the points where its derivatives are zero or undefined. Since $\frac{\partial g}{\partial x}$ is undefined when $x=0$, it follows that $\frac{\partial \mathcal{L}}{\partial x}$ is undefined at $x=0$ and that the points $(0,1)$ and $(0,-1)$ need to be examined (plus the other two boundary points with $y=0$).

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  • $\begingroup$ Could you explain how this criteria applies to the set $S_2$ in my question that has no boundary points, nor why $S_1$ does have boundary points ? $\endgroup$
    – me10240
    May 27 '19 at 7:58
  • $\begingroup$ I understand now, I was fixated on the fact that the points happened to be end points of the curve described by $S_1$. I see now that the equation $\nabla f = \lambda \nabla g$ has meaning , and is solved, only for points where $\nabla f, \nabla g$ make sense. Clearly, the points where $\nabla g$ are undefined will need to be added to the list of extrema candidates. $\endgroup$
    – me10240
    May 30 '19 at 1:56

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