Lagrange Multipliers and Inequalities

**fobos3** · July 22nd 2010, 03:19 PM

Is there a way to apply the method of Lagrangian Multipliers when your equation of constraint is an inequality(it won't be called an equation of constraint then I guess). Here is an example of what I came up with.

Say $f(x)=x^2$

You want to find the extrema satisfying the inequality $-1\le x\le1$

If you introduce a variable $z$ that satisfies the equation $x^2+z^2=1$ that means $x\in[-1,1]$

$g(x,z)=x^2+z^2=1$ becomes your equation of constraint. Now we have

$\nable f=\lambda\nabla g$ where $\nabla =\left(\dfrac{\partial}{\partial x},\dfrac{\partial}{\partial z}\right)$

We get $\left(2x,0\right)=\lambda\left(2x,2z\right)$

From $2\lambda z=0$ we get $z=0$ or $\lambda=0$

From $2x=2\lambda x$ we get $x=0$ or $\lambda=1$

For $x=0$ we get our minimum.

For $\lambda=1$ we have $z=0$ and $x^2=1\therefore x=\pm 1$ which is the local maximum.

Now what I want to know is, if there is a way to solve for more general inequalities like $x<3$ or $x>7$ .

Any input will be appreciated.

**Media_Man** · July 24th 2010, 06:38 AM

In principle, I think the answer is yes, by introducing the simple disk as you have, which was a good idea. So for $x<3$ , you would let $g(x,z)=x^2+z^2=9$ . For $x>7$ , the domain is a bit trickier, but you can replace with $\frac1x<\frac17$ and make $g(x,z)=\frac1{x^2}+z^2=\frac1{49}$ to get $x>7$ or $x<-7$ . I think this might be what you are looking for.

However, keep in mind that in one dimension, there is little point to Lagrange multipliers. Consider: if $\nabla f = \lambda \nabla g$ , then by letting $\lambda=0$ , $\nabla f=0$ , so $(\frac{\partial}{\partial x}, 0)=(0,0)$ , i.e. $f'(x)=0$ , which is the standard one-dimensional way of finding the maxima/minima anyway, having nothing to do with $g(x,z)$ .

You might try experimenting with your method on functions $f(x,y)$ over two variables to see if there is something truly new to be discovered.

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