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$f(x) = x-y$ subject to constraint $x^2-y^2=2$
Using the method of Lagrange Multipliers, we get:

$(1, -1) = \lambda(2x, -2y)$ which gives $x=y$ but this does not satisfy $x^2-y^2=2$

Is this because the Lagrange method is not applicable here? Or is it valid for the Lagrange method to not produce any solutions?

EggHead
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1 Answers1

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Lagrange multipliers work in the following situation.

Suppose $g(x,y)=0$ defines a bounded level curve. Suppose also that $\nabla g \neq 0$, that is the gradient is non-zero, on this level set. Then the max an min of $f$ on $g=0$ occurs at a point that satisfies $\nabla f = \lambda \nabla g$ and $g=0$.

What if you just assume that $g(x,y)=0$ and allow for $\nabla g=0$? Then the min and max of a $f$ will occur at a point where $\nabla g =0 $ or where $\nabla f = \lambda \nabla g$ in addition to $g=0$.

However, your problem is that your constraint does not define a bounded set. Think about it, if you have an unbounded set and you are trying to maximize the distance to the origin, is this possible? No. This is the sort of situation you have.

abnry
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  • true, g(x,y) = 0 defines a hyperbola which is unbounded. Does that mean that we have to use the "Hessian" insteade of Lagrange in this case? – EggHead Nov 08 '13 at 22:09
  • No, the answer is that no min or max exist. Draw the curve $x^2-y^2=2$ and draw the level sets of $f(x,y)=x-y$, which are the lines $y=x+c$. You'll see that for $c$ as large as you like will intersection $x^2-y^2=2$ somewhere. Same for $c$ as negative as you like. – abnry Nov 09 '13 at 00:04
  • thanks. I figured it out right after I posted the last comment. The book had a little note (not a Theorem!) that mentioned that if the constraining surface was unbounded like a hyperboloid then f need not have a maxima or minima. And in this case $z= x^2-y^2-2$ is indeed a hyperboloid. That was very instructive... Thanks again! – EggHead Nov 09 '13 at 03:12