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Homework Help: 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).

    So:
    [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

    HallsofIvy

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    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 )
     
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