# Lagrange multiplier problem

1. Apr 23, 2005

### Pepsi24chevy

Hey guys, i need some help with this problem. It goes as follow: Find the global max and min valves of the fuction z=x^2+2y^2 on the circle x^2+y^2=1. Ok and here is what i have done. I found the derivites and have done (K is the lagrange constant)
2x=2xK K=1
4y=2yK K=2

Then i set them equal and get 1=2? I know this is not right, but my teacher said do it by lagrange multipliers, so any help is appreciated. Thanks.

2. Apr 23, 2005

### arildno

Why do you assume that both x and y must be different from zero?

3. Apr 23, 2005

### dextercioby

$$\{x=0;y=\pm 1\} \mbox{and} \{y=0;x=\pm 1\}$$ should be the solution...I think the second delivers the minimum value...

Daniel.

4. Apr 23, 2005

### Pepsi24chevy

From looking at it I can see how you get those values but how would i do this mathamatically? WOuld i just take 1=2=0? Then go back and plug is or what. I am just trying to understand this, thanks for helping me out.

5. Apr 23, 2005

### snoble

Just substitute $$x^2+y^2 =1$$ into $$z= x^2+ y^2+y^2= 1+ y^2$$. Then just take the derivative of this in terms of y. Once you know the y value where the maximum happens just solve for x such that $$x^2= 1- y^2$$

Oh and since you are on the unit circle you also need to check the bound $$y=\pm 1$$

Last edited: Apr 23, 2005
6. Apr 23, 2005

### Pepsi24chevy

I got to do this by lagrange multipliers though, that is what is messing me up because I don't see how it can be done by using this method.

7. Apr 23, 2005

### arildno

Pepschevy:
You've got 3 equations in 3 unknowns:

$$2x=2xK, 4y=2yK, x^{2}+y^{2}=1$$
Agreed?
Your flaw has been to divide the first equation with 2x, and the second equation with 2y.
This is only possible to do at the same time if neither x or y is 0!!

But, furthermore:
You have proven that if we assume this we get 1=K=2, that is, 1=2 a contradiction.

Thus, you must conclude:
Either x or y MUST be zero!

Last edited: Apr 23, 2005