# Lagrange multipliers - confused about when the constraint set has boundary points that need to be considered

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 $$S_1$$, $$S_2$$.

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.

• Does this mean that if the level set is not a closed curve, then I will have a boundary point ? May 27, 2019 at 8:46
• 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. May 27, 2019 at 8:57
• 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 May 27, 2019 at 9:06

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 $$y=0$$).

• 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 ? May 27, 2019 at 7:58
• 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. May 30, 2019 at 1:56