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
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