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My book tells me that of the solutions to the Lagrange system, the smallest is the minimum of the function given the constraint and the largest is the maximum given that one actually exists.

But what if we only have one point as a solution? How to know whether Lagrange multipliers gives maximum or minimum?

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    $\begingroup$ If your constraints describe a closed and bounded domain (that is, a bounded domain with a boundary), then we must attain both a maximum and a minimum. So, it's impossible to get only one critical point. $\endgroup$ – Ben Grossmann Dec 30 '16 at 1:31
  • $\begingroup$ @Omnomnomnom However, if the region is not compact, it is possible to get only one critical point. Examples can be found here, here and here. $\endgroup$ – Pedro Dec 30 '16 at 4:12
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    $\begingroup$ I have updated my answer as it was incorrect. Please see the new one and see the comments on it for details. $\endgroup$ – user307169 Dec 30 '16 at 4:52
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    $\begingroup$ Note that in Lagrange multipliers theorem, you assume that the maximum/minimum exists on the set of constraints and the method only gives candidates for it. Consider for example $f(x,y)=x+y^3$ along the $y$-axis (this is, $g(x,y)=x=0$. The gradient of the functions are parallel at the origin but $f$ has no maximum and no minimum on the $y$-axis. $\endgroup$ – Taladris Dec 30 '16 at 9:28
  • $\begingroup$ @Taladris Do I understand correctly that the Langrange method does not always give even local extrema? (The way I read the theorem is that the extrema satisfies the condition - but other points may also?) $\endgroup$ – Tony Jul 10 '20 at 1:23
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As Om(nom)$^3$ said in the comments, if you're working on a closed and bounded region then it's not possible to get only one critical point.

If you're not on a closed and bounded region then it's no longer guaranteed that you'll have more than one critical point. If you only have one critical point then you can use the Bordered Hessian technique. (Thanks to ziggurism for clearing that up.)

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  • $\begingroup$ Are you sure the normal multivariable second derivative test applies? I was taught a different technique for constrained problems, the bordered Hessian en.m.wikipedia.org/wiki/Hessian_matrix#Bordered_Hessian $\endgroup$ – ziggurism Dec 30 '16 at 1:55
  • $\begingroup$ @ziggurism is that for all constrained problems or just constrained problems on a bounded region? The wiki was unclear unless I glossed over it. $\endgroup$ – user307169 Dec 30 '16 at 2:09
  • $\begingroup$ being a local extremum is a local property. I can't see how it can be affected by the boundedness of the region. Should I try to concoct an example where normal second derivative says max, but bordered hessian says min? $\endgroup$ – ziggurism Dec 30 '16 at 2:24
  • $\begingroup$ @ziggurism I'd be curious to see one. I'm mainly just wondering what the criteria are for the bordered hessian to apply since the first sentence in that section says it's for "certain" constrained problems, implying it doesn't work in all constrained problems. Unboundedness of the region is just a guess on my part since it's one of the ways a region will be no longer compact. $\endgroup$ – user307169 Dec 30 '16 at 2:30
  • $\begingroup$ I'm not sure what they meant with "certain constrained systems", but my guess would be holonomic constraints, i.e. exactly those constrained systems to which the method of Lagrange multipliers applies. $\endgroup$ – ziggurism Dec 30 '16 at 2:32
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On a closed bounded region a continuous function achieves a maximum and minimum. If you use Lagrange multipliers on a sufficiently smooth function and find only one critical point, then your function is constant because the theory of Lagrange multipliers tells you that the largest value at a critical point is the max of your function, and the smallest value at a critical point is the min of your function. Thus max = min, i.e. the function is constant. Also note that "critical point" should probably be called something else, like "point of interest" because usually critical points are defined as points where the gradient is zero.

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  • $\begingroup$ Consider $f(x,y)=x^2+y^2$ with $x+y=1$ Lagrange gives $<2x,2y>=\lambda<1,1>$ this gives $x=\frac{1}{2}=y$ and $\lambda=1$. My function is not constant, so what do you mean by on a sufficiently smooth function. $\endgroup$ – Ahmed S. Attaalla Dec 30 '16 at 5:20
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    $\begingroup$ @AhmedS.Attaalla your function is not restricted to a closed bounded region and thus my statement does not apply. $\endgroup$ – nullUser Dec 30 '16 at 5:44
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In fact the normal second derivative test doesn't apply to constrained extremum problems. You should instead use the Bordered Hessian method. In brief, instead of computing the positive-definiteness of the Hessian matrix of second partial derivatives of $f$, you instead compute the Hessian of $f-\lambda g$, including derivatives with respect to $\lambda$

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In the Lagrange Multipliers Method the points obtained will be critical points (solutions of an equation which have the form $\nabla f(x)=\lambda\varphi(x)$) of an objective function $f$ (of class $C^1$) restrict to a region $M$ which have the form $M=\varphi^{-1}(c)$, where $\varphi$ is a function (of classe $C^1$) that comes from the constraint (which have the form $\varphi(x)=c$).

Usually, the existence of the maximum and the minimum comes from the continuity of $f$ and the compactness of $\overline{M}$. In this case, $f$ have at least two critical points on $\overline{M}$. However, there are cases in which the equation $\nabla f(x)=\lambda\varphi(x)$ give us only one solution $p\in M$ (this is because the other critical point is in $\overline{M}\setminus M$). Here is a possible approach that sometimes works for these cases:

  • Show that the maximum (or minimum) is not in $\overline{M}\setminus M$.

  • Conclude that $p$ is the maximum (or minimum) of $f$ on $M$.

You can see examples of this case here, here and here.

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You could also just pick another point satisfying the constraints, evaluate the function for that point, and see if that value is higher or lower than what you found with Lagrange multipliers.

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    $\begingroup$ That only works if the extreme value is absolute. It won't necessarily work if it's a relative extreme value. $\endgroup$ – user307169 Dec 30 '16 at 2:08
  • $\begingroup$ But if there's only one extreme point, wouldn't that work? $\endgroup$ – Nathan H. Dec 30 '16 at 3:15
  • $\begingroup$ you'd also need to make sure it's actually an extreme point and not a saddle point. $\endgroup$ – user307169 Dec 30 '16 at 3:37
  • $\begingroup$ Yeah, you're right. Just read on Wikipedia that the points you get are possible maxima/minima, but they aren't necessarily maxima/minima. I forgot about that. $\endgroup$ – Nathan H. Dec 30 '16 at 3:49

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