Max/Min of f Using Lagrange Multipliers

[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?

 

[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?

 

[alejandrito29]

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?

thank, but, i don't understand :(

 

[arildno]

What is it you don't understand??

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

 

[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 ?

 

[HallsofIvy]

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?

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.

 

[alejandrito29]

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.

Very Thanks