Question

use lagrange multipliers to find the extreme values of f(x,y)=x^2-y^2 subject to the constraint x^2+y^2=1. I) the extreme values of f(x,y), subject to said constraint, occur at those points (a,b) on R^2 such that <2a,-2b>=lambda<2a,2b> for some lambda on R and a^2+b^2=1. true or false?

Answer 1

i answered true. im pretty sure it fits the first criteria. just not sure about the second. i assume the second one is by extension true

Answer 2

im not sure about the (a,b), sounds true
if you work it out though the extreme value of f(x,y) =-1 for x=0, y=+-1

Answer 3

or y=0, x=+-1 works too

Answer 4

does there need to be one value of lambda that satisfies the lagrange equation?

I wrote out the lagrange equation and got:

\[\Lambda(x,y,\lambda)=x^2-y^2+\lambda(x^2+y^2-1)\]
taking partials:

\[\Lambda_{x}=2x+2x \lambda=0, \Lambda_{y}=-2y+2y \lambda=0\]
so for the first partial lambda must be negative, and for the second lambda must be positive. So there isn't a single multiplier so that the gradient of the lagrange equation is zero.

Answer 5

i agree with it being true. the second part has three answer choices.
a) the extrema occur at (+-1,0) and (0,+-1)
b) the extreme values are -1 and 1
c) all of the above

Answer 6

I geuss I need to study this more

Answer 7

thats what i got
A) then

Answer 8

what about the extreme values?

Answer 9

oops yeah so all of the above

Answer 10

ok good. I feel better about this one too. I really appreciate the help.

Answer 11

rsvitale, i think its ok if multiplier can be either 1 or -1

Answer 12

allright so we can have a seperate multiplier for x and y

Answer 13

as long as solution fits all three equations

Answer 14

allright

Answer 15

i appreciate the help from both of you.

Answer 16

your welcome, good luck with rest of your classes

Answer 17

thanks