Results 1 to 13 of 13

Math Help - Please help - Stationary points

  1. #1
    Member
    Joined
    Jul 2008
    Posts
    212

    Please help - Stationary points

    Find the stationary points of the following functions, and
    identify the nature of each (ie local maxima, minima, saddle point or point of inflexion):

    (a) f(x, y) = x2 + y2 - x - y

    (b) g(x, y) = 3x2 - 3xy + 4y2 + 8x - 5y - 4


    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,
    Quote Originally Posted by brumby_3 View Post
    Find the stationary points of the following functions, and identify the nature of each (ie local maxima, minima, saddle point or point of inflexion):
    (a) f(x, y) = x2 + y2 - x - y
    (b) g(x, y) = 3x2 - 3xy + 4y2 + 8x - 5y - 4
    Find x_0 and y_0 such that \frac{\partial f}{\partial x}(x_0,y_0)=0 and \frac{\partial f}{\partial y}(x_0,y_0)=0

    These are the critical points.


    I hope you've done bits of algebra including matrices...

    The Hessian matrix is defined at a point as :

    H_f(x_0,y_0)=\begin{pmatrix} \frac{\partial^2 f}{\partial x^2}(x_0,y_0) & \frac{\partial^2 f}{\partial x \partial y}(x_0,y_0) \\ \\ \frac{\partial^2 f}{\partial y \partial x}(x_0,y_0) & \frac{\partial^2 f}{\partial y^2}(x_0,y_0) \end{pmatrix}

    \frac{\partial^2 f}{\partial x^2}=\frac{\partial}{\partial x} \left(\frac{\partial f}{\partial x}\right) \quad \quad \quad \quad \frac{\partial^2 f}{\partial x \partial y}=\frac{\partial}{\partial x} \left(\frac{\partial f}{\partial y}\right)

    Work with eigenvalues
    If the eigenvalues are all positive, then this point (x_0,y_0) is a maximum.
    If they are all negative, then this point (x_0,y_0) is a minimum.
    If one is positive and one is negative, then this point (x_0,y_0) is a saddle point.

    Work with discriminant
    If the discriminant is negative, then it's a saddle point.
    If the discriminant is positive and \frac{\partial^2 f}{\partial x^2}(x_0,y_0) >0, then it's a maximum.
    If the discriminant is positive and \frac{\partial^2 f}{\partial x^2}(x_0,y_0) <0, then it's a minimum.

    If you don't know how to proceed, google for Hessian matrix.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by brumby_3 View Post
    Find the stationary points of the following functions, and
    identify the nature of each (ie local maxima, minima, saddle point or point of inflexion):

    (a) f(x, y) = x2 + y2 - x - y

    (b) g(x, y) = 3x2 - 3xy + 4y2 + 8x - 5y - 4


    First find where \frac{\partial f}{\partial x}=0 and \frac{\partial f}{\partial y}=0 [and in part (b), where the partials of g are equal to zero.]

    These will give you the critical points. (Be careful on how you pair them together)

    Then apply the second partials test to see if its a max, min, or is a saddle point.

    D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-f_{xy}^2(x_0,y_0)

    Recall that if:

    D>0~\text{and}~f_{xx}(x_0,y_0)>0, then f has a relative minimum at (x_0,y_0)

    D>0~\text{and}~f_{xx}(x_0,y_0)<0, then f has a relative maximum at (x_0,y_0)

    D<0, then f has a saddle point at (x_0,y_0)

    D=0, then no conclusion can be drawn.

    You have enough information to help you get started on the problem(s)

    --Chris
    Last edited by Chris L T521; September 14th 2008 at 12:01 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jul 2008
    Posts
    212
    I've never heard of the Hessian matrix and it doesn't seem to be mentioned in my course outline. Can someone please go through this question step by step as an example so that I can try and do the next homework questions myself?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jul 2008
    Posts
    212

    Stationary points

    Beginning of merged thread...
    Last edited by Jameson; September 30th 2008 at 03:33 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    I'm fairly certain that points of inflection are not considered stationary points. That being said, given f(x, y), you must calculate all points at which df/dx = df/dy = 0, and then you must calculate f_{xx}, f_{xy}, f_{yx}, f_{yy} and evaluate f_{xx}f_{yy} - f_{xy}f_{yx}. If that value is positive and both f_{xx} and f_{yy} are positive, it means that the point is a minimum. If that value is positive and both f_{xx} and f_{yy} are negative, it means that the point is a maximum. If f_{xx}f_{yy} - f_{xy}f_{yx} is negative, the point is a saddle point. Edited for correctness.
    Last edited by icemanfan; September 19th 2008 at 02:22 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jul 2008
    Posts
    212
    Can you show how to do it using the questions asked please?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    (a): \frac{df}{dx} = 2x - 1; \frac{df}{dy} = 2y - 1; f_{xx} = 2; f_{yy} = 2; f_{xy} = 0; f_{yx} = 0

    Hence we have x = 0.5; y = 0.5 and thus our only point of consideration is (0.5, 0.5). What type of point is this? Use the formula:

    f_{xx}f_{yy} - f_{xy}f_{yx} = 2 \cdot 2 - 0 \cdot 0 = 4 - 0 = 4

    Since 4 is positive and both 2 and 2 are positive, this point is a local minimum. In this case, it is also an absolute minimum of the function.

    See if you can handle (b).

    Edited for clarity and correctness.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    If you need an example, we need only do one of these, right?

    the rules are, if (a,b) is a critical point of a function f(x,y), that is, f_x(a,b) = f_y(a,b) = 0 (and of course we assume f is differentiable on some disk containing (a,b) and has continuous partial derivatives on the said disk etc), then we can classify the critical points as follows:

    Let D = D(a,b) = f_{xx}(a,b)f_{yy}(a,b) - [f_{xy}]^2, then

    (a) If D > 0 and f_{xx}>0, then f(a,b) is a local minimum

    (b) If D > 0 and f_{xx}<0, then f(a,b) is a local maximum

    (c) If D<0, then f(a,b) is a saddle point.


    lets see how this works with your f(x,y)

    Note that:

    \begin{array}{lcl} f_x = 2x - 1 & & f_y = 2y - 1 \\ & & \\ f_{xx} = 2 & & f_{yy} = 2 \\ & f_{xy} = 0 & \end{array}

    so that x = y = \frac 12 are the only values of x and y that make our partials zero. hence, the only critical point is \bigg( \frac 12 , \frac 12 \bigg)

    Now, D \bigg( \frac 12 , \frac 12 \bigg) = 4 > 0 and f_{xx} \bigg( \frac 12 , \frac 12 \bigg) = 2 > 0, so that f \bigg( \frac 12 , \frac 12 \bigg) is a local minimum


    EDIT: I should have known iceman would pick up the slack
    Last edited by Jhevon; September 19th 2008 at 02:32 PM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Jul 2008
    Posts
    212
    Okay I've looked at b) and I'm still not sure how to go around it? Am I suppose to collect like terms first?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by brumby_3 View Post
    Okay I've looked at b) and I'm still not sure how to go around it? Am I suppose to collect like terms first?
    collect like terms where? where are you stuck? you found the partial derivatives already, right? what did you get?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Jul 2008
    Posts
    212
    lol I'm lost altogether
    Follow Math Help Forum on Facebook and Google+

  13. #13
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by brumby_3 View Post
    lol I'm lost altogether
    you must be able to at least find the partial derivatives. if not, then us going through these examples will not help you.

    please review finding partial derivatives.

    to start you off: when you find the derivative of a function with respect to one variable, you treat all other variables as constants. until you can do this, it makes no sense to try these problems. you are essentially trying to run before you even know how to crawl
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: August 24th 2011, 11:35 AM
  2. stationary points
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 16th 2009, 05:00 AM
  3. Stationary Points
    Posted in the Calculus Forum
    Replies: 4
    Last Post: December 3rd 2008, 12:18 PM
  4. Stationary Points, Local points of Extrema
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 19th 2008, 01:44 PM
  5. Replies: 3
    Last Post: May 5th 2006, 09:22 AM

Search Tags


/mathhelpforum @mathhelpforum