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Constrained extrema

  1. Nov 14, 2011 #1
    1. The problem statement, all variables and given/known data

    find the local extrema of

    f(x,y) = 6x^2 -8x + 2y^2 - 5 around the closed disk x^2 +y^2 =< 1

    2. Relevant equations

    3. The attempt at a solution

    I used the larangian way of doing this but not sure if its right. My solution:

    F(x,y) = f(x,y) + lambda g(x,y)

    where g(x,y) = x^2 + y^2 - 1

    taking partials wrt to x y and lambda I get

    Fx = 12x - 8 + 2x lamda

    Fy = 4y + 2y lamda

    Flamda = g(x,y)

    setting each to 0 and solving I get

    lamda = 2 from Fy

    subbing into Fx I get x = 1

    subbing into g(x,y) and solving I get y = ±1

    so the extrema are

    (1,1) (1,-1)

    edit: I think I shouldve had my equation as

    F(x,y) = f(x,y) - lamda g(x,y) rather than f(x,y) + lamda g(x,y)

    with that equation I get

    x = 0.5
    y = ±sqrt 0.75
    Last edited: Nov 14, 2011
  2. jcsd
  3. Nov 14, 2011 #2

    Ray Vickson

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    When you have an inequality constraint g <= 0, your Lagrangian should be [itex] F - \lambda g [/itex] (with [itex] \lambda \geq 0[/itex]) for a _maximization_ problem and should be [itex] F + \lambda g [/itex] (with [itex] \lambda \geq 0[/itex]) for a _minimization_ problem. Alternatively, you could use the single form [itex] F - \lambda g [/itex] for both max or min problems, but now [itex] \lambda \geq 0[/itex] for a max problem and [itex] \lambda \leq 0[/itex] for a min problem.

    Basically, the way to remember this is to say that the Lagrangian should be *better than* the objective F for feasible points, so if feasible points have g <= 0 we get something bigger than F by subtracting a positive constant times g (and bigger is better when maximizing), and we get something smaller than F by adding a positive constant times g, and smaller is better when minimizing.

  4. Nov 14, 2011 #3
    so in this case, where the critical points are supposed to fall within that disk, does that mean x and y will have to both be less than the radius of the disk (ie any points in the disk)?
  5. Nov 15, 2011 #4


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    The Lagrange multiplier method will help you find extrema on the circumference of the disk. To find extrema in the disk, simply set the partial dervatives of the object function to 0 and see if you find in critical points in the disk.
  6. Nov 15, 2011 #5

    Ray Vickson

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    I don't understand your question. In your original post you said you were confused about whether you should write F + lambga*g or F - lambda*g, as these gave different solutions. All I did was answer _that_ question; I did not say anything about x and y being less than the radius of the disk or equal to the radius, or anything else like that.

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