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Using lagrange mutilpliers when lamda has two values

  1. May 7, 2012 #1
    Consider $${f (x, y) = x^2 + 2 y^2}$$ subject to the constraint $${x^2 + y^2 = 1}$$.
    What would be the minimun and the maximum values of the f.

    Trouble is when I tried solving the problem lamda comes out to have two values 1 and 2 respectively. How do I proceed in order to get the answers?
  2. jcsd
  3. May 7, 2012 #2

    Two points:

    1) You get two different values for lambda only if you assume [itex]\,\,xy\neq 0\,\,[/itex] . Drop this assumption.

    2) Why Lagrange, anyway? You can do way easier, imo: [tex]x^2+y^2=1\Longrightarrow x^2=1-y^2\Longrightarrow f(x,y)=(1-y^2)+2y^2=y^2+1[/tex] and this last is an easy to check 1-variable function...!

  4. May 7, 2012 #3
    Thing is that I'm required to solve it using Lagrange multipliers.

    Now, I'm confused about you saying that drop the assumption xy≠0.

    Could you do me a favour and show your working. I reached the point where we get two values of lamda but after that am clueless.
  5. May 7, 2012 #4

    Some people around here think I answer too much, or do homework, for some people. In your case I feel the same since you've

    been asking a lot lately...

    Anyway, I can show you a little, I guess: if you define [itex]\,\,H(x,y,\lambda):= x^2+2y^2-\lambda(x^2+y^2-1)\,\,[/itex] and calculate the partial derivatives, you get:

    [itex]H_x=0\Longleftrightarrow x=\lambda x\Longrightarrow \,\,[/itex] either [itex]\,\,x=0\,,\,\,\,any\,\, \lambda\,\,[/itex] , or else [itex]\,\,x\neq 0\,\,,\,\,\lambda=1[/itex] ...

    The same, of course, is for for the partial in y and etc.

  6. May 8, 2012 #5
    The reason I'm asking a lot lately is that I've got my calculus-II final exam coming up in a few days and currently am more than a little worried about it. Anyway thanks about the post.

    P.s. "Some people around here think I answer too much, or do homework, for some people."

    Just so you know, I sincerely appreciate all that you do.
  7. May 12, 2012 #6


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    Any value of [itex]\lambda[/itex] is not relevant to the solution and you should not even bother to evaluate it! The way I would do this is note that
    [tex]\nabla x^2+ 2y^2= \lambda\nabla x^2+ y^2[/tex]
    [tex]2x\vec{i}+ 4y\vec{j}= 2\lambda x\vec{i}+ 2\lambda y\vec{j}[/tex]
    so that we have the two equations
    [tex]2x= 2\lambda x[/tex]
    [tex]4y= 2\lambda y[/tex]

    Because a specific value of [itex]\lambda[/itex] is not a part of the solution, eliminate it first by dividing the first equation by the second:
    [tex]\frac{2x}{4y}= \frac{2\lambda x}{2\lambda y}[/tex]
    [tex]\frac{x}{2y}= \frac{x}{y}[/tex]

    xy= 2xy which is equivalent to xy= 0. Either x= 0 or y= 0. If x= 0, [itex]x^2+ y^2= y^2= 1[/itex] so [itex]y= \pm 1[/itex]. If y= 0, [itex]x^2= 1[/itex] so [itex]x=\pm 1[/itex]. The four solutions are (0, 1), (0, -1), (1, 0), and (-1, 0).
  8. May 12, 2012 #7


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    Homework Helper

    The division is only valid in the first place if [itex]y \neq 0[/itex], so the y = 0 solution doesn't follow from that method directly. One then either has to divide the y equation by the x equation to find that or just work with the original system of equations.
  9. May 13, 2012 #8


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    You just have to check that each solution satisfies the original constraint, and all four do. The two x=0 solutions are equal maximums and the two y=0 solutions are equal minimums. This is an easy problem to check visually if you can picture it.
  10. Jun 26, 2012 #9
    am pretty new to lamdba calculus and would need some help.
    would anyone be able to help me on how to find the linked and free variables in the following terms..
    M1 = lambda x.(f(x x))
    M2 = lambda f.(M1 M1)
    M3 = (M2 lambda y.x)

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