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Find the maximum and minimum values of the function (Answers included).

  1. Mar 26, 2012 #1

    s3a

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    1. The problem statement, all variables and given/known data
    Find the maximum and minimum values of the function
    f(x,y) = 2x^2 - 28xy + 2y^2 - 1 on the disk x^2 + y^2 <= 1.

    (a) Find the maximum.
    (b) Find the minimum.


    2. Relevant equations
    Lagrange Multiplier method and partial differentiation.


    3. The attempt at a solution
    My work is attached as MyWork.jpg. If anything is unclear, tell me and I will restate it (or whatever it is you request). In the image, I forgot to plug in x and y in f(x,y) to get f(sqrt(1/2), sqrt(1/2). This yields the correct answer for the minimum: 2*(sqrt(1/2))^2 - 28*sqrt(1/2)*sqrt(1/2) + 2(sqrt(1/2))^2 - 1 whereas the correct answer for the maximum is: 2*(sqrt(1/2))^2 + 28*sqrt(1/2)*sqrt(1/2) + 2(sqrt(1/2))^2 - 1. So, my ultimate question is, what are the reasons for choosing positive or negative square roots for x and/or y?
     

    Attached Files:

  2. jcsd
  3. Mar 26, 2012 #2

    Ray Vickson

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    Homework Helper

    In the problem max f(x) s.t. g(x) <= 0, if you want the Max you should use L = f - λg, with λ ≥ 0; if you want the Min you should use L = f + λg, with λ >= 0 (or use f - λg with λ ≤ 0).

    How can you remember this? The easiest way is to form L so that it is more favourable than f for feasible points; that is, for feasible points, L should be ≥ f in a max problem and should be ≤ f in a min problem. So, if the constraint is g ≤ 0 we get something ≥ f in a max problem by subtracting g, that is, by using L = f - λg with λ ≥ 0.

    Note that for an equality constrained problem, with constraint g = 0, the sign of λ is not determined, and it does not matter whether you write L = f + λg or L = f - λg.

    So, in your analysis you need to look at the sign of your λ.

    RGV
     
  4. Mar 26, 2012 #3

    HallsofIvy

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    The Lagrange multiplier method requires that you have "=" in the constraint. You can use that to determine any max or min on the boundary. Just set the partial derivatives equal to 0 in the interior of the disc.
     
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