# Lagrange multiplier

1. Jun 20, 2011

### alejandrito29

In a exercise says:

Find max a min of $$f=-x^2+y^2$$ abaut the ellipse $$x^2+4y^2=4$$

i tried $$-2x=\lambda 2x$$
$$2y=\lambda 8y$$
$$x^2+4y^2-4=0$$

then $$\lambda =-1$$ or $$\lambda =\frac{1}{4}$$ , but, ¿how i find $$x,y$$????

2. Jun 20, 2011

### arildno

Well, suppose you identically fulfill your first eqaution by choosing lambda=-1.

What does this imply that "y" must equal, by looking at your second equation?

3. Jun 20, 2011

### alejandrito29

thank, but, i dont understand :(

4. Jun 20, 2011

### arildno

What is it you don't understand??

What does the second equation look like if you insert lambda=-1?

5. Jun 20, 2011

### Skins

Okay you seem to be going along fine. From your first equation you determined that $$\lambda = -1$$ So now if you plug $$\lambda = -1$$ into your second equation what must y be ? and when you plug that into your third equation what do you get for x ? Now from your second equation you determined $$\lambda = \frac{1}{4}$$ so when you plug that into your first equation what must x be ? and then what do you get for y when you plug into your third equation ?

6. Jun 20, 2011

### HallsofIvy

If $\lambda= -1$ then the second equation becomes $2y= -8y$ so that $y= 0$. You can solve $$x^2= 4$$ for the corresponding x values.

If $\lambda= \frac{1}{4}$, then the first equation becomes $-2x= (1/2)x$ so that $x= 0$. Solve $4y^2= 4$ for y.

7. Jun 20, 2011

Very Thanks