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Global max/min (multivariable)

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the global maximum and global minimum of
    [tex]f(x, y)=x^3+y^2[/tex]
    in the half disk [tex]x^2+y^2\leq{1}, y\geq{0}[/tex]

    This is a rather new topic for me, so I am looking for clarification for problems such as these.

    3. The attempt at a solution
    First, I determined the critical points of the function.
    [tex]\nabla{f(x, y)=3x^2i+2yj}[/tex] and thus has a critical point at
    [tex](x, y)=(0, 0)[/tex]

    Next, we will consider along the boundary of the half disk. Since the disk is a half circle, I must include the endpoints x=1, x=-1.

    We have,
    [tex]x=cost, y=sint, (0, \pi{)}[/tex]
    We have,
    [tex]f(cost, sint)=cos^3t+sin^2t=g(t)[/tex]
    Moreover,
    [tex]-3cos^2tsint+2sintcost=0\rightarrow{cost=2/3}[/tex]
    [tex]g(cos^{-1}2/3)=0.852[/tex]
    Next, checking the boundaries at t=0 and t=π
    g(0)=1, g(π)=-1

    Therefore, f has a global max at the boundary on the half disk of 1 at (1, 0) and a global min of -1 at (-1, 0).

    However, I executed two of the tests for determining if critical points are max/mins for the point (0. 0) however all the tests were inconclusive. What else can be done here?
     
  2. jcsd
  3. Nov 5, 2011 #2

    SammyS

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    You checked along the boundary, x2 + y2 = 1, but to be thorough, you should also check along the boundary, y = 0 .

    Regarding (0,0): Are the second derivatives of f defined at (0,0) ?
     
  4. Nov 5, 2011 #3
    Okay, not entirely sure how to show work for checking along y=0. Moreover, I believe the second derivatives are defined at (0, 0), and if they are not, I don't see why considering it's a cubic and square function.
     
  5. Nov 5, 2011 #4

    I like Serena

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    Hi sandy.bridge! :smile:

    In (0,0) you'll find that the Hessian 2nd derivative test is inconclusive.
    What else can be done?
    Well, since there are no other extrema nearby, simply check a few points around (0,0).
    Since it's already likely that it is a saddle point, you can verify this by finding a positive and negative value near (0,0).

    As for checking along y=0.
    What about [itex]x=t, y=0[/itex]?
    In particular you will find that (0,0) is a saddle point.
     
  6. Nov 5, 2011 #5
    Hello!
    Okay, moving the left of the origin along the x-axis, we have
    [tex]f(-.1, 0)=(-.1)^3[/tex]
    and ro the right
    [tex]f(.1, 0)=0.1^3[/tex]
    therefore, this illuminates the fact this is a saddle point.
    As for your suggestion of x=t, how can this be done? Is "t" not denoting the angle measurement with respect to the positive x-axis?
     
  7. Nov 6, 2011 #6

    I like Serena

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    Good!

    "t" can be anything you want. It's just an arbitrary parameter with an arbitrary range.
    You could also use simply (x,0) where x is the parameter running from -1 to +1.
     
  8. Nov 6, 2011 #7
    Alright, then the global max occurs when x=1, or t=0(angle). Also, the min occurs when x=-1 (t=pi).
    max(x, y)=(1, 0)
    min(x ,y)=(-1, 0)
     
  9. Nov 6, 2011 #8

    I like Serena

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    And... we're done! :cool:
     
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