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Lagrange Multipliers HELP!

  1. May 6, 2008 #1
    We're suppose to minimize f(x,y,z)=x^2+y^2+z^2 subject to 2x+y+2z=9.

    I only ever remember learning how to do f(x,y) would it be the same equation? Thus, f(x,y,[tex]\lambda[/tex]) = f(x,y) + [tex]\lambda[/tex] g(x,y)? Meaning f(x,y,z,[tex]\lambda[/tex]) = x^2+y^2+z^2 + [tex]\lambda[/tex] (2x+y+2z-9) and then continue solving for each variable from there?

    Any help?
  2. jcsd
  3. May 6, 2008 #2


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    "Lagrange Multipliers" doesn't apply to a single equation.

    "Lagrange Multipliers" says that at a min or max of f(x,y,z), subject to the additional condition that g(x,y,z)= constant, the two gradients must be parallel: one is a multiple of the other. [itex]\nabla f= \lambda \nabla g[/itex]. That is the same as saying [itex]\nabla \left(f(x,y,z)+ \lambda g(x,y,z)\right)= 0[/itex] for all [itex]\lambda[/itex] s what you suggest will work.

    Please do not post the same thing more than once. I have merged your two posts.
    Last edited by a moderator: May 6, 2008
  4. May 6, 2008 #3
    so you're saying that if i use f(x,y,z,[tex]\lambda[/tex]) = x^2+y^2+z^2 + [tex]\lambda[/tex] (2x+y+2z-9) that i will be able to find the value of each after taking partial derivatives of each variable. So would I just have three points?

    p.s. - Sorry!!
  5. May 6, 2008 #4


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    Yes, take the gradient of f, set it equal to 0 and the three partial derivatives give you three equations to solve for x, y, z (eliminating [itex]\lambda[/itex], reduces that to 2 equations but you also know 2x+ 2y+ 2z= 9.)

    I'm not sure what you mean by "just have three points". There is one point on that plane that minimizes f.
  6. May 6, 2008 #5
    Yea, I misspoke. I meant one point with three numbers meaning a x-value, y-value, and z-value.
  7. May 6, 2008 #6
    what do you mean eliminating [itex]\lambda[/itex] and reducing to two equations.

    f_x = 2x+2[itex]\lambda[/itex]
    f_y = 2y+[itex]\lambda[/itex]
    f_z = 2z+2[itex]\lambda[/itex]
    2_[itex]\lambda[/itex] = 2x+y+2z-9

    and from there find the value of each and put them back into f(x,y,z)=x^2+y^+z^2
  8. May 7, 2008 #7


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    YOu haven't set them equal to 0 yet! If [itex]2x+ 2\lambda= 0[/itex] and [itex]2z+ 2\lambda= 0[/itex], what do you get when you eliminate [itex]\lambda[/itex]?
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