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Lagrange Multiplier question with solid attempt!

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Use the method of Lagrange multipliers to find the maximum and minimum values of the function

    f(x, y) = x + y2

    subject to the constraint g(x,y) = 2x2 + y2 - 1


    2. Relevant equations

    none

    3. The attempt at a solution

    We need to find [itex]\nabla[/itex]f = λ[itex]\nabla[/itex]g

    Hence,

    [itex]\nabla[/itex]fx - λ[itex]\nabla[/itex]gx = 0

    Which becomes, 1 - λ(4x) = 0

    [itex]\nabla[/itex]fy - λ[itex]\nabla[/itex]gy = 0

    Which becomes, 2y - λ = 0

    -------------------------

    Now we have: x = 1/4λ and y = λ/2

    I assume I am right in now subbing x and y into the constraint....

    To give us: 2/16λ2 + λ2/4 - 1 = 0


    It seems a bit messy considering this is an elementary part of my homework? Have I gone wrong somewhere?

    Regards as always
     
  2. jcsd
  3. Nov 3, 2011 #2

    Ray Vickson

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    Your [itex] \partial{L}/\partial{y}[/itex] is incorrect; try again.

    RGV
     
  4. Nov 3, 2011 #3
    Wow, rookie error..

    yep dL/dy = 2y - λ(2y)
    so λ = 1

    Thus, x=1/4

    Put this into constraint to get: y=7/8

    So (1/4,7/8) is a point..

    How do i find the maximum and minimum? thanks
     
  5. Nov 3, 2011 #4

    Ray Vickson

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    the equation 2y = lambda*(2y) has TWO solutions: lambda = 1 is one of them. Can you see another?

    RGV
     
  6. Nov 3, 2011 #5

    HallsofIvy

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    Since specific values of [itex]\lambda[/itex] are not part of the "solution" of problems like these, I find it is often a good step to start by dividing one equation by another, eliminating [itex]\lambda[/itex] at the start. Here, your equations are [itex]1= 4\lambda x[/itex] and [itex]2y= 2\lambda y[/i2tex]. Dividing the second equation by the first:
    [tex]\frac{2y}{1}= \frac{2\lambda y}{\lambda x}= \frac{2y}{x}[/tex]
    which reduces to x= 1 or y= 0.

    However, you say your constraint is "[itex]g(x,y)= 2x^2+ y^2- 1[/itex]" which is not a constraint at all. Did you mean [itex]g(x,y)= 2x^2+ y^2= 1[/itex]? If so, set x= 1 in that and solve for y, then set y= 0 and solve for x.
     
  7. Nov 4, 2011 #6

    Ray Vickson

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    The statement that "specific values of lambda are not part of the solution ..." is either misleading or wrong, depending on exactly how you meant it. ONE of the max or min points IS obtained by looking at lambda = 1 and seeing the consequences.

    RGV
     
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