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Global max/min on disc

  1. Jul 11, 2014 #1
    1. The problem statement, all variables and given/known data

    find the global extrema on the disc [tex] x^2 + y^2 \le 1[/tex]

    given the function [tex]f(x,y)=xy+5y[/tex]




    3. The attempt at a solution

    For the interior of the disc

    [tex]\nabla f = <y,x+5>[/tex]

    the critical point is (0,-5)

    for the boundary of the disc
    using lagrange multipliers

    [tex] \left\{\begin{array}{cc}y=\lambda 2x \\ x+5 = \lambda 2y \\ x^2+y^2 =1 \end{array}\right.[/tex]

    solving for lambda

    [tex] \lambda = \frac{y}{2x}; \lambda = \frac{x+5}{2y} [/tex]
    [tex]\frac{y}{2x}=\frac{x+5}{2y} \Rightarrow y = \pm \sqrt{x(x+5)}[/tex]
    now,
    [tex]x^2 + (\pm \sqrt{x(x+5)})^2 = 1 \Rightarrow x = \frac{1}{4}(\pm\sqrt{33}-5)[/tex]
    subbing the x value into
    [tex]y = \pm \sqrt{x(x+5)} \Rightarrow \pm \sqrt{\frac{1}{8}(5 \sqrt{33} - 21)}[/tex]

    I know to test those critical points in the original function but before I go further I want to make sure I have done everything up to it correctly
     
    Last edited: Jul 11, 2014
  2. jcsd
  3. Jul 11, 2014 #2

    LCKurtz

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    I got the same numbers using a different method.
     
  4. Jul 12, 2014 #3

    Ray Vickson

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    Be careful: one of your ##x =\frac{1}{4}(\sqrt{33}-5)## and ## x =\frac{1}{4}(-\sqrt{33}-5)## is incorrect; do you see why?

    It would have been less troublesome to use the Lagrangian stationary conditions to solve for ##x,y## as functions of ##\lambda##, then use these expressions in the constraint to get a single equation for ##\lambda##. The two roots for ##\lambda## correspond to the min and the max, and when you then use those values in the ##x,y## expressions you are done: no ##\pm## problems to worry about.
     
  5. Jul 12, 2014 #4
    I am not currently aware of Lagrangian stationary conditions. I will use my googles but I always welcome better ways to do a problems. especially ones with less than optimal results.

    I'll check those x values in the morning.

    EDIT:

    I found something about creating a new function and taking partials

    [tex]F(x,y,z,\lambda)=f(x,y,z) - \lambda(g(x,y,z)-k)[/tex]

    and take the partials for x,y,z,lambda
     
    Last edited: Jul 12, 2014
  6. Jul 12, 2014 #5

    Ray Vickson

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    You are aware of the conditions: you used them! They are just the two equations ##x+5 = 2 \lambda y## and ##y = 2 \lambda x##. These are the "stationarity" conditions for the Lagrangian function ##L = f(x,y) - \lambda g(x,y)## in the problem
    [tex] \min f(x,y)\\
    \text{subject to } g(x,y) = 0[/tex]

    BTW: the Lagrangian ##L## plays an important role in second-order tests for maxima and minima, but more on that later.
     
  7. Jul 12, 2014 #6
    ah, okay thank you. I just hadn't heard that terminology used yet.
     
  8. Jul 12, 2014 #7
    I went through it again, got the same number. I am not seeing why one of the x values is incorrect
     
  9. Jul 13, 2014 #8

    Ray Vickson

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    Whenever you use nonlinear manipulations (squaring equations, and the like) you can introduce false solutions, that is, so-called solutions that do not satisfy the original equations you started with. Check both ##x = (1/4)(\sqrt{33}-5)## and ##x = (1/4)(-\sqrt{33}-5)##. There is no way that one of them can possibly satisfy ##x^2 + y^2 \leq 1## for any real value of ##y##. If you don't believe it, try evaluating them numerically.
     
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