1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lagrange Multiplier!

  1. Nov 5, 2005 #1
    Hi, I would appreciate if anyone can help me out with the following question.

    I've been asked to find the point on the surface z = xy + 1 nearest to the origin by using the Lagrange Multiplier method. But all the examples I've been given in class and for coursework gave you the constraint equation.

    Is there a constraint equation given in this question? :confused:
  2. jcsd
  3. Nov 5, 2005 #2
    Well it's not given in the sense of, constraint equation = ?
    But you should be able to set one up as long as you know the distance formula.
    [tex] d = \sqrt{(z_2-z_1)^2+(x_2-x_1)^2+(y_2-y_1)^2} [/tex]
    x, and y are going to be arbitrary and you have the expression for z. You also know that you are aiming for the orgin, so [tex] \vec 0 = (x_1,y_1,z_1) [/tex].
  4. Nov 5, 2005 #3
    Actually 'shoot'.. that doesn't make any sense

    Your constraint would be that 'the point MUST be on that surface' (that sounds more like a constraint to me).

    [tex] g(x,y,z) = z-xy=1 [/tex]
    [tex] f(x,y,z) = d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2} [/tex]
    Where [tex](x_1,y_1,z_1)=\vec 0 [/tex]

    Ok, I think that makes sense now...
  5. Nov 5, 2005 #4
    Hmm sorry I'm still lost :(
    Last edited: Nov 5, 2005
  6. Nov 5, 2005 #5
    sorry I'm not sure how to use Latex, but what are the values for x2, y2 and z2? :confused: O god, I don't understand this at all :uhh:
  7. Nov 5, 2005 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    The constraint is z = xy + 1!! If you were asked simply to "find the point closest to the origin" then the answer would be (0,0) itself. But you are not asked that- you are asked to fine the point on z= xy closest to the origin.
    Using the Lagrange multiplier method: The square of the distance of a point (x,y) to (0,0) (minimizing the square of distance is the same as minimizing distance itself) is minimizing x2+ y2. The gradient of that is 2xi+ 2yj. The gradient of the constraint, z- xy= 1, is -yi- xj+ k. Lagrange's method says that one of those must be a multiple of the other:
    2xi+ 2yj= k(-yi- xj+ k) which tells us 2x= -ky, 2y= -kx, and 0= k. What x,y,z satisfy those?
  8. Nov 6, 2005 #7
    hmm ok...I got x = 0, y = 0 and z = 1...is that right? :confused: so the point is ( 0 , 0 , 1 )
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Lagrange Multiplier!
  1. Lagrange Multiplier (Replies: 2)

  2. Lagrange Multipliers (Replies: 7)

  3. Lagrange Multipliers (Replies: 8)

  4. Lagrange Multipliers (Replies: 8)