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Lagrange multipliers (yes, again)

  1. May 5, 2014 #1
    1. The problem statement, all variables and given/known data
    agasfasd.png

    f=xy^2

    C: x^2 + y^2 = 3

    2. Relevant equations



    3. The attempt at a solution

    I don't understand how he can say that x=0 is a solution in this one. Looking at the contours, there are no solutions for f if x=0.
     
  2. jcsd
  3. May 6, 2014 #2

    Ray Vickson

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    I also don't understand it. In fact, it is false. Points of the form ##(0 \pm \sqrt{3})## are neither maxima nor minima of f, because as ##x## increases through 0, f increases from negative to positive if ##y^2 = 3##: it looks like ##f = 3x## for fixed ## y = \pm \sqrt{3}##---that is, if we go along a line parallel to the x-axis. Since such a line is tangent to the constraint circle, the same type of behavior will happen if we go a small distance along the circle instead of its tangent line. In other words, points with x = 0 are neither local maxima nor minima in the constrained problem.

    Anyway, points with x = 0 are not among those we get from the Lagrange multiplier method.
     
  4. May 6, 2014 #3
    I thought it seemed suspicious, though I have to admit I didn't fully understand your explanation.
     
  5. May 6, 2014 #4

    Ray Vickson

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    Think about it, take your time, draw pictures, plug numbers into formulas, etc.
     
  6. May 6, 2014 #5

    BvU

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    Function is not f = xy^2 but f = yx^2. Print is barely readable but teachers handwriting confirms.
    I agree with teacher!
     
  7. May 6, 2014 #6

    HallsofIvy

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    Langrange's method gives the two equations [itex]2xy= 2\lambda x[/itex] and [itex]x^2= 2\lambda y[/itex]. What I usually do in such a case is eliminate [itex]\lambda[/itex] by dividing one equation by the other, say dividing the first equation by the second, to get [itex]2xy/x^2= x/y[/itex] but since we cannot divide by 0, that is only valid if x and y are not 0. We need to check those separately. If x= 0, the first equation, [itex]2xy= 2\lambda x[/itex], is satisfied for all y. The second equation, [itex]x^2= 2\lambda y[/itex] is satisfied if y= 0 or [itex]\lambda= 0[/itex]. If both x and y are 0, the equation [itex]x^2+ y^2= 3[/itex] cannot be satisfied so that is NOT a solution. If x and [itex]\lambda[/itex] are 0, the equation [itex]x^2+ y^2= 3[/itex] becomes [itex]y^2= 3[/itex] so [itex](0, \sqrt{3})[/itex] and [itex](0, -\sqrt{3})[/itex] are valid solutions.

    If y= 0, the equation [itex]x^2= \lambda y= 0[/itex] is satisfied only for x= 0 and we have already seen that (0, 0) does not satisfy [itex]x^2+ y^2= 3[/itex].
     
  8. May 6, 2014 #7

    Ray Vickson

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    I use the OP's (wrong) description that f = xy^2. I will not bother to look at the OP's submitted messy and illegible photocopies, and since the OP cannot manage to submit correct problem descriptions (on more than one occasion) I will no longer respond to his postings.
     
    Last edited: May 6, 2014
  9. May 7, 2014 #8
    Well excuse me for being human and making mistakes. Burn me at the the stake, right?

    Thanks HallsOfIvy that's a great explanation!
     
  10. May 7, 2014 #9

    BvU

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    No need to get burnt (both). I do think that if you manage to antagonize Ray by apparently repeating human errors, there is something there for you to pick up: at least you could promise to try to correctly formulate problem statements (*1) and check if you don't make it difficult for potential helpers (*2). I was on the wrong track for a while too.

    (*1) it is human to make that kind of mistake and then overseeing subsequent pointers that it's the other way around. But you can learn from it.
    (*2) different matter, more about attitude, meticulousness etc. From that one you can learn that investing a little time to do it completely right saves time and goodwill in the long run

    All in good, constructive spirit

    Personally I am a slow learner and I do repeat mistakes quite often. Hence the (moderate amount of) understanding sympathy.
     
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