Apparently the following general statement is true.

"Let $\gamma:g(x,y)=0$ be a closed curve that doesn't cross itself. If the maximisation of a function $f(x,y)$ on $g(x,y)$ using Lagrange multipliers gives three solutions, then one of the 3 is an absolute maximum, the other is an absolute minimum and the third is a flat region (saddle point?) of the function $f\vert_\gamma$".

This statement seems far too general to be true. Can't all of the solutions for example be local maxima or minima?


I think I can give you an intuitive argument. Say you parametrize $\gamma$ by a function $\vec r(t)$ and then consider $f(\vec r(t))$. If the Lagrange method gives you three stationary points, then you have: $$\left.\frac{\partial f(\vec r(t))}{\partial t}\right|_{t=t_i} = 0$$for $i=1,2,3$; i.e. $f$ has stationary points on $\gamma$ in the points corresponding to $\vec r(t_1)$, $\vec r(t_2)$ and $\vec r(t_3)$. Consider a sign table for the first derivative of $f$ on $\gamma$:

$$\begin{array}{c|cccccccc} t & & \color{blue}{t_1} & & t_2 & & t_3 & & \color{blue}{t_1} \\ \hline f(\vec r (t)) & & \color{blue}{f(\vec r (t_1))} & & f(\vec r (t_2)) & & f(\vec r (t_3)) & & \color{blue}{f(\vec r (t_1))} \\ f'(\vec r (t)) & \color{blue}{?} & \color{blue}{0} & ? & 0 & ? & 0 & \color{blue}{?} & \color{blue}{0}\\ \end{array}$$

Since $\gamma$ is closed, I extended the table to emphasize that the blue parts on the left and right ends of the table have to be equal (overlapping part; the table is "circular"). The four question marks are to be replaced with the possible signs of $f'(\vec r (t))$.

WLOG we can assume $\color{blue}{?}$ to be positive:

$$\begin{array}{c|cccccccc} t & & \color{blue}{t_1} & & t_2 & & t_3 & & \color{blue}{t_1} \\ \hline f(\vec r (t)) & & \color{blue}{f(\vec r (t_1))} & & f(\vec r (t_2)) & & f(\vec r (t_3)) & & \color{blue}{f(\vec r (t_1))} \\ f'(\vec r (t)) & \color{blue}{+} & \color{blue}{0} & \color{red}{?} & 0 & \color{red}{?} & 0 & \color{blue}{+} & \color{blue}{0}\\ \end{array}$$

The possibilities for the two remaining $\color{red}{?}$'s are:

  1. $+$ , $+$
  2. $+$ , $-$
  3. $-$ , $+$
  4. $-$ , $-$

Note that case 1 is impossible since then $f'(\vec r (t)) \ge 0$ on the entire parameter domain, in conflict with the closed path $\gamma$. Cases 2, 3 and 4 all correspond to the situation of 1 maximum, 1 minimum and 1 inflection point.

  • $\begingroup$ Thank you for this detailed explanation! The only thing I don't understand is that I thought that lagrange multipliers can only give you maxima and minima. Can they also in fact give you saddle points?? $\endgroup$ – john melon Jun 14 '16 at 15:12
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    $\begingroup$ The method of Lagrange's multipliers gives you critical/stationary points, the minima and maxima are among those. $\endgroup$ – StackTD Jun 14 '16 at 15:22
  • $\begingroup$ I am still persuaded that Lagrange's multipliers only give extrema, thus the exclusion of saddle points. Am I missing something? $\endgroup$ – john melon Jun 14 '16 at 15:34
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    $\begingroup$ Constraining $z=f(x,y)$ to the curve $\gamma$ given as $g(x,y)=0$ gives you a curve (in 3D), this can have minima, maxima and inflection points, not saddle points. But in general; check this example to see it doesn't necessarily give you an extremum. $\endgroup$ – StackTD Jun 14 '16 at 15:38
  • $\begingroup$ OK. It is clear now $\endgroup$ – john melon Jun 14 '16 at 15:58

If you accept without proof that such a curve has a regular parametric representation $$t\mapsto{\bf z}(t)=\bigl(x(t),y(t)\bigr)\qquad(t\in{\mathbb R}/{\mathbb Z})$$ of period $1$ the claim can be proven as follows: Consider the pullback $$\phi(t):=f\bigl({\bf z}(t)\bigr)\ .$$ Convince yourself by means of the chain rule that $\phi'(t_0)=0$ iff ${\bf z}(t_0)$ is one of the points produced by Lagrange's method. Therefore we know now that $\phi$ has exactly three critical points per period, and is strictly monotone in between these. We need one critical point $t_1$ for the global maximum of $\phi$ and a second critical point $t_2$ for the global minimum. If, e.g., $t_1<t_2<t_3<t_1+1$ then $\phi$ would be increasing immediately to the left of $t_3$ as well as immediately to the right of $t_3$. This excludes a local extremum of $\phi$ at $t_3$.

  • $\begingroup$ Thank you! I assume this is why it is not stated that $t_3$ is a saddle point? $\endgroup$ – john melon Jun 14 '16 at 15:13
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    $\begingroup$ Saddle points occur only on manifolds of dimension $\geq2$. $\endgroup$ – Christian Blatter Jun 14 '16 at 15:22

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