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Lagrange multiplier

  1. Jun 20, 2011 #1
    In a exercise says:

    Find max a min of [tex]f=-x^2+y^2[/tex] abaut the ellipse [tex]x^2+4y^2=4[/tex]

    i tried [tex]-2x=\lambda 2x[/tex]
    [tex] 2y=\lambda 8y [/tex]
    [tex]x^2+4y^2-4=0[/tex]

    then [tex]\lambda =-1[/tex] or [tex]\lambda =\frac{1}{4}[/tex] , but, ¿how i find [tex]x,y[/tex]????
     
  2. jcsd
  3. Jun 20, 2011 #2

    arildno

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    Well, suppose you identically fulfill your first eqaution by choosing lambda=-1.

    What does this imply that "y" must equal, by looking at your second equation?
     
  4. Jun 20, 2011 #3
    thank, but, i dont understand :(
     
  5. Jun 20, 2011 #4

    arildno

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    What is it you don't understand??

    What does the second equation look like if you insert lambda=-1?
     
  6. Jun 20, 2011 #5
    Okay you seem to be going along fine. From your first equation you determined that [tex]\lambda = -1[/tex] So now if you plug [tex]\lambda = -1[/tex] into your second equation what must y be ? and when you plug that into your third equation what do you get for x ? Now from your second equation you determined [tex]\lambda = \frac{1}{4}[/tex] so when you plug that into your first equation what must x be ? and then what do you get for y when you plug into your third equation ?
     
  7. Jun 20, 2011 #6

    HallsofIvy

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    If [itex]\lambda= -1[/itex] then the second equation becomes [itex]2y= -8y[/itex] so that [itex]y= 0[/itex]. You can solve [tex]x^2= 4[/tex] for the corresponding x values.

    If [itex]\lambda= \frac{1}{4}[/itex], then the first equation becomes [itex]-2x= (1/2)x[/itex] so that [itex]x= 0[/itex]. Solve [itex]4y^2= 4[/itex] for y.
     
  8. Jun 20, 2011 #7
    Very Thanks
     
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