2
$\begingroup$

I am new with Lagrange multipliers , and having trouble understanding what is a necessary condition and what is sufficient.

Assume I want to find global exterma of $f(x,y,z) \quad s.t. \quad g(x,y,z)=0$. As far as I understand, Lagrange multipliers is a necessary condition (right?), so my question is: Is it possible that the function $f$ has no exterma under the constraint $g=0$ (It does not have any maximum or minimum under this constraint ), but when I use Lagrange multiplies, I will obtain solutions (i.e.- points where $\nabla L =0$ ) ?

Hope I made myself clear

Thanks a lot in advance

$\endgroup$
0
0
$\begingroup$

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\grad}{\nabla}$Let $f$ and $g$ be continuously-differentiable real-valued functions on $\Reals^{3}$, and assume $\grad g$ is non-vanishing on the (non-empty) level set $\{g = 0\}$, so that $M = \{g = 0\}$ is a smooth surface by the implicit function theorem.

The Lagrange multipliers equation $$ \grad f = \lambda \grad g \tag{1} $$ is the critical point condition for $f$ on $M$. As you say, this is necessary in order that $f$ have an extremum, but it is not generally sufficient, as you suspect.

To give a simple example, if $g(x, y, z) = z$, then $M = \{g = 0\}$ is the $(x, y)$-plane, and a solution of (1) is precisely a critical point of the function $\phi(x, y) = f(x, y, 0)$. It's easy to cook up functions $f$ so that $\phi$ has no extrema (say, $f(x, y, z) = x^{2} - y^{2}$), or has absolute extrema but also critical points that are not absolute extrema.

If $M$ is compact, i.e., closed and bounded, then the extreme value theorem guarantees that $f$ achieves an absolute minimum and an absolute maximum on $M$. Nonetheless, $f$ can still have critical points on $M$ that are not extrema. The prototypical example is perhaps to pick real numbers $a < b < c$, and to consider $$ f(x, y, z) = ax^{2} + by^{2} + cz^{2},\quad g(x, y, z) = x^{2} + y^{2} + z^{2} - 1. $$ The solutions of (1) are easily checked to be the standard basis vectors and their negatives; the absolute minima of $f$ occur at $\pm(1, 0, 0)$, the absolute maxima at $\pm(0, 0, 1)$. The points $\pm(0, 1, 0)$ are non-extremal critical points.

$\endgroup$
2
  • $\begingroup$ Thanks for your answer. Just to make sure, assume we are dealing with the two constraints: $2x^2+y^2-4=0 , \quad x+y+z=0$. If I wouldn't know that these two constraints give actually an ellipse, which is a compact set in $\mathbb{R}^3$, I wouldn't have any other way to prove that a global maximum or minimum exists? $\endgroup$ Jan 13 '16 at 1:30
  • $\begingroup$ Generally, showing a continuous function has absolute extrema, to say nothing of finding the extrema, is Difficult. That's one reason we put so much effort into compactness, why we spend so much time on critical points in calculus (this can reduce one's search from an uncountable set to a finite set!), why calculus of variations is hard, etc. It may be possible to show a function has absolute extrema without compactness ($f(x) = x^{2}$ on $(-1, 1)$, say), but one has to expect to work harder, and to meet failure more frequently. (Minimize $|\zeta(x + iy)|^{2}$ if $x \geq -0.4999$...!) $\endgroup$ Jan 13 '16 at 2:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.