# Lagrange theorem help

As per Lagrange theorem it says I can get maximum or minimum under some constraint.
Example:

$f(x,y)=x^2+y^2+4$ under constraint $x+y=2$. I can use Lagrange theorem for this.

But I have a problem like $f(x,y)=x^2+y^2+4$ under constraint like $x$ should be minimum.

How can I solve this any theorem or solution?

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Is the equation $2x$ or $x^2$? – EuYu Oct 14 '12 at 19:34
x(square) I just showed example.I am interested in knowing which theorem to solve this? – constantlearner Oct 14 '12 at 19:36
What do you mean by saying "$x$ should be minimum"? If you're optimizing both $f$ and $x$, you have to say something about the relative importance of these objectives; otherwise the problem isn't fully defined. See also en.wikipedia.org/wiki/Multi-objective_optimization. – joriki Oct 14 '12 at 19:39
What do you mean by $x$ should be minimum? – EuYu Oct 14 '12 at 19:39
When you say the constraint can be ambiguous, that normally poses some sort of problem. In your problem, you can make $x$ as "minimal" as you want by taking $x\rightarrow -\infty$. But what exactly are you trying to accomplish? – EuYu Oct 14 '12 at 19:45

Use Lagrange multipliers with the constraint $\partial g/\partial x=0$. This will at least constrain $g$ to be extremal. So construct an auxillary function

$$F(x,y,\lambda)=f(x,y)+\lambda \frac{\partial g(x)}{\partial x}$$

and minimize with $\nabla F=0$

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I doubt if a general solution for multiple optimalization exists, since the goals might be just conflicting with each other. But I assume you must observed that the function $$x^{2}+y^{2}+4$$ has an absolute minimum at $x=0,y=0$. On the other hand if you let $x=0$, $f(0,y)=y^{2}+4$ has no maximum at all. So there is essentially only one choice.

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Use $∇f = -\lambda∇g$. $\lambda$ is called the Lagrange multiplier.

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