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Lagrange multiplier question

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the maximum and minimum values of f(x,y) = x5y3 on the circle defined by x2 + y2 = 10. Do the same for the disc x2 + y2 ≤ 10.


    3. The attempt at a solution
    for the first part, if I call the circle g(x,y) defined by x2 + y2 = 10
    I need to now define some F(x,y,λ) = f(x,y) - λg(x,y) and find the critical points of this, i.e.: where Fx', Fy', and Fλ' = 0. I was wondering if I should include the constant (10) in the function λg, or does it matter at all? i.e.:
    F(x,y,λ) = x5y3 - λx2 - λy2 + 10λ
    OR
    F(x,y,λ) = x5y3 - λx2 - λy2

    (from there I can figure it out) I was just unclear as to how the constraint is handled.
     
  2. jcsd
  3. Nov 5, 2011 #2
    I doubt think the constant makes a difference as you'll have to differentiate g,f.
     
  4. Nov 5, 2011 #3
    yes but it matters when you differentiate for λ
     
  5. Nov 5, 2011 #4

    BruceW

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    Differentiating with respect to [itex]\lambda[/itex] gives you the equation you already know [itex]x^2 + y^2 = 10[/itex]. So for the sake of completeness, you should keep the 10 in the equation.

    But if you're not interested in differentiating with respect to [itex]\lambda[/itex] (because you already know what it will tell you), then you could just take the 10 out.
     
  6. Nov 5, 2011 #5
    alright thank you!
     
  7. Nov 5, 2011 #6

    Ray Vickson

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    The _standard_ Lagrangian approach to max/min f(x1,x2,...,xn) subject to g(x1,x2,..,xn) = 0 is to look at L = f +- u*g (using u instead of lambda) and to set the partials dL/d x_j = 0 for j = 1,...,n. So, in your case, g = x^2 + y^2 - 10; that is, the '10' is part of the definition of g. Of course it goes away when you take xj-derivatives. Keeping the '10' is certainly necessary if you set dL/du = 0.

    Why does this matter? Well, if you use the convention L = f - u*g, and look at a 'nearby' problem max/min f subject to g = a (for small |a|) then the optimal value function F(a) (= optimal value of f, subject to g(x) = a) satisfies dF/da|_{a=0} = u, where u = the Lagrange multiplier in the a=0 problem. So, for example, suppose you wanted the maximum value (not the x,y that give the maximum) for the related problem x^2 + y^2 = 10.1. This corresponds to g = 0.1, because the '10' is part of the definition of g. The optimal value in the new problem would be (approx) the old optimal value + 0.1*u .

    RGV
     
  8. Nov 5, 2011 #7
    But for the second part, the disk defined by x2 + y2 ≤ 10,
    How is this any different? at least in this case, the max and min are on the outer circle so there is no max or min when x and y are less. Is there a different method of solving this?
     
  9. Nov 5, 2011 #8

    Matterwave

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    I believe for the second part all you have to show is that the max occurs on the boundary and so the answer should be the same as part 1. Extremize f, and show that there is only a minimum at some point inside the boundary, and then use the extreme-value theorem to show that the maximum must be on the boundary (i.e. a maximum must exist).
     
  10. Nov 5, 2011 #9

    Ray Vickson

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    You need to worry about whether or not there is an interior optimum (where x^2 + y^2 < 10); at such a point the Lagrange multiplier = 0 and you are just looking at stationary points of f(x,y) itself. In this case a direct argument is easiest: given any interior point (x0,y0), argue that there are other points (x1,y1) and (x2,y2) giving f(x1,x1) > f(x0,y0) > f(x2,y2); you actually need to carry out that argument, but it is not hard.

    Note: in this case it is important that x and y not have sign restrictions (so negative x and/or y are OK). If you also had restrictions x >= 0 and y >= 0, you would need to be more careful.

    RGV
     
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