Results 1 to 3 of 3

Math Help - Minimum point of f(x,y)

  1. #1
    Newbie
    Joined
    Jun 2007
    Posts
    15

    Minimum point of f(x,y)

    I have the following function: f(x,y)=x^4+2x^3y+x^2y^2+y^4 and I know that there is a critical point at (0,0). I have been asked to determined the type of critical point. The second derivatives give no information, and I'm not supposed to consider higher derivatives.


    "Obviously" it is a minimum point, so I thought it would be enough to show that f(x,y)\geq 0 near (0,0).

    By introducing polar coordinates I got:

    f(r\cos(\theta),r\sin(\theta))=...=r^4\left[1+2\cos^3(\theta)\sin(\theta)-cos^2(\theta)\sin^2(\theta)\right].

    Since r^4\geq 0, it should be enough to show that 1+2\cos^3(\theta)\sin(\theta)-cos^2(\theta)\sin^2(\theta)\geq 0, but I seem unable to do it. Does this seem like the right method, and if so, does anyone know how to go on from here?


    Another thought I had was to compare with (x+y)^4 or (x-y)^4, but that didn't get me anywhere.


    Aliquantus
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,368
    Thanks
    43
    Quote Originally Posted by Aliquantus View Post
    I have the following function: f(x,y)=x^4+2x^3y+x^2y^2+y^4 and I know that there is a critical point at (0,0). I have been asked to determined the type of critical point. The second derivatives give no information, and I'm not supposed to consider higher derivatives.


    "Obviously" it is a minimum point, so I thought it would be enough to show that f(x,y)\geq 0 near (0,0).

    By introducing polar coordinates I got:

    f(r\cos(\theta),r\sin(\theta))=...=r^4\left[1+2\cos^3(\theta)\sin(\theta)-cos^2(\theta)\sin^2(\theta)\right].

    Since r^4\geq 0, it should be enough to show that 1+2\cos^3(\theta)\sin(\theta)-cos^2(\theta)\sin^2(\theta)\geq 0, but I seem unable to do it. Does this seem like the right method, and if so, does anyone know how to go on from here?


    Another thought I had was to compare with (x+y)^4 or (x-y)^4, but that didn't get me anywhere.


    Aliquantus
    Note the following

    x^4 + 2x^3y+x^2y^2+y^4 = x^2\left(x^2+2xy+y^2\right) + y^4 = x^2\left(x+y\right)^2+y^4.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2007
    Posts
    15
    Of course, thanks alot!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: February 26th 2011, 01:12 PM
  2. Minimum point on a quadratic curve
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 4th 2010, 10:47 AM
  3. maximum and minimum point
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 6th 2009, 07:12 AM
  4. gradient, minimum point
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 23rd 2009, 12:52 PM
  5. Minimum Point
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 21st 2009, 09:23 AM

Search Tags


/mathhelpforum @mathhelpforum