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Lagrange Multipliers with ellipse

  1. Jul 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the points on the ellipse x2 + 2y2 = 1 where f(x,y) = xy has its extreme values.

    2. Relevant equations



    3. The attempt at a solution
    f(x,y,z) = x2 + y2 + z2 -- constraint
    g(x,y,z) = x2 + 2y2 -1 = 0

    gradient of f = [tex]\lambda[/tex] * gradient of g

    2xi + 2yj + 2zk = [tex]\lambda[/tex]2xi + [tex]\lambda[/tex]4yj

    2x = [tex]\lambda[/tex]2x
    2y = [tex]\lambda[/tex]4y
    2z = 0

    I don't know where to go from here, can someone help me out?
     
  2. jcsd
  3. Jul 22, 2010 #2
    Sorry, I made a mistake in my original post, it should be:

    f(x,y,z) = xy
    g(x,y,z) = x2 + 2y2 - 1

    yi + xj = [tex]\lambda[/tex](2xi + 4yj)
    y = 2x[tex]\lambda[/tex]
    x = 4y[tex]\lambda[/tex]

    y = 2(4y[tex]\lambda[/tex])[tex]\lambda[/tex] = 8y[tex]\lambda[/tex]2

    [tex]\lambda[/tex] = sqrt(1/8)

    where do i go from here?
     
  4. Jul 22, 2010 #3

    Office_Shredder

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    you have three equations and three unknowns.... the two gradient equations and the fact that x2+2y2=1. Now that you have solved for one unknown, can you make it two equations with two unknowns?
     
  5. Jul 22, 2010 #4
    How about this one:

    Find the minimum distance from the surface x2 - y2 - z2 = 1 to the origin. (function being minimized = x2 + y2 + z2)

    2xi + 2yj + 2zk = [tex]\lambda[/tex](2xi - 2yj -2zk)

    2x = 2x[tex]\lambda[/tex]
    2y = -2y[tex]\lambda[/tex]
    2z = -2z[tex]\lambda[/tex]
    x2 = y2 +z2 + 1

    - I can't figure out how to solve this system of equations, any tips?
     
  6. Jul 23, 2010 #5

    HallsofIvy

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    These are pretty close to being trivial. From the first equation, if [itex]x\ne 0[/itex], [itex]\lambda[/itex] must be 1. From the second equation, if [itex]y\ne 0[/itex], [itex]\lambda[/itex] must be -1, and from the third, if [itex]z\ne 0[/itex], [itex]\lambda[/itex] must be -1.

    Since [itex]\lambda[/itex] cannot be both 1 and -1, at least one of the coordinates must be 0!

    Suppose x= 0. Then the constraint becomes [itex]-y^2- z^2= 1[/itex] which is impossible. If x is not 0, then [itex]\lambda[/itex] must be 1, not -1, so y and z must be 0. The constraint becomes [itex]x^2= 1[/itex].

    Looking at it geometrically gives an easy check. This is a "hyperboloid of two sheets" with the x-axis as axis of symmetry.
     
    Last edited: Jul 23, 2010
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