Question

Lagrange Multipliers:

I'm having an issue with the process of lagrange multipliers. I know that the equations are:
1. g(x,y) = 0
2. fx = lambda gx
3. fy = lambda gy
For some reason though, in this particular problem I get 2 different values for lambda so I'm kinda stuck. HELP! I'll attach the problem in the comments.

Answer 1

Here's my work:
Constraint: 4x^2 + y^2 - 4 = 0
What to minimize/maximize: Distance formula sqrt(x^2 + y^2)
To make things easier, we can max/min the square of the distance formula instead.

Answer 2

I hate these things
I found the problem in the other one btw, if you want to check that post
the notifications aren't working...

Answer 3

oh haha I'll take a look

Answer 4

ok that makes sense. any idea where to go with this one?

Answer 5

df/dx = 2x, df/dy = 2y, dg/dx = 8x, dg/dy = 2y

Answer 6

totally have to review LM's, so I'll need a minute
I should eat first too....

Answer 7

hm.... sounds like a good idea. Where you at?

Answer 8

Guadalajara Mexico, you?

Answer 9

Chicago.

Answer 10

right on, I"m originally from Cali
what's your major?

Answer 11

Undecided Engineering. Leaning Mechanical or Electrical. Studying at NU

Answer 12

cool, I'm probably going CS and Electrical Engineering in Minneapolis
alright going to eat, see you in a bit!

Answer 13

ok I'll be here trying to figure these things out.

Answer 14

ok I'm ready to help, are you there?

Answer 15

yup

Answer 16

first of all. g(x,y)=k
your partials are right, so by\[\nabla f(x,y,z)=\lambda\nabla g(x,y,z)\\g(x,y,z)=k\]we get the system\[8x=2\lambda x\\2y=2\lambda y\\4x^2+y^2=4\]we only need to solve for \(\lambda\) once, it doesn'tmatter that you get two different values for it if you try to solve with each equation (you actually can avoid finding lambda all together in some problems, but don't think in this one)

Answer 17

so use the first equation to solve for \(\lambda\) and \(x\)

Answer 18

From the first equation we get that lambda is 1/4, but how does that help us?

Answer 19

because the x's cancel...

Answer 20

lambda is 4...
we can also solve for x, too

Answer 21

oh right. Lambda is 4. but how can we solve for x if it cancels out of the equation?

Answer 22

\[8x-2\lambda x=0\implies x=0,~\lambda=4\]

Answer 23

OHHH

Answer 24

it's the only possible solution...

Answer 25

so plug that value for x into the constraint and solve for y to get one set of critical points

Answer 26

so if x = 0 y = 2, -2 right?

Answer 27

right :)
now we can use the second equation and \(\lambda=4\) to solve for y

Answer 28

then plug that back into the constraint, solve for x, and get another set of critical points

Answer 29

there we get that lambda is 1 and y is 0 right?

Answer 30

and if y = 0 , x = 1,-1

Answer 31

this is where lambda doesn't matter, if you use lambda=4 you get the same answer ;)

Answer 32

:P

Answer 33

that's why I say you can avoid solving for lambda at all sometimes
funny little trick mr Lagrange came up with lol

Answer 34

so anyway....our critical points are (0,2), (0,-2), (1,0), (-1,0) and we just plug those into the original equation to figure out which are max's and which are min's. nice. thanks! you saved my rear end again....

Answer 35

no problem, you gave me a review again :) a fair trade in my book!

Answer 36

All's well that ends well.