Results 1 to 4 of 4

Math Help - Stationary points / two variables (help)

  1. #1
    Member billym's Avatar
    Joined
    Feb 2008
    Posts
    183

    Stationary points / two variables (help)

    I have this function and I am looking for the stationary points:

    <br />
f(x,y) = x^4 + 2x^2y^2 + y^4 + 4x^2 - 4y^2 + 6<br />

    I started by trying to solve the two partial derivatives:

    <br />
4x(x^2 + y^2 + 1) = 0,  (E.1)<br />
    <br />
4y(y^2 + x^2 - 1) = 0,  (E.2)<br />

    If x = 0, then (E.2) becomes:

    <br />
4y(y^2 - 1) = 0<br />

    Thus y = 0 , y = 1 , or y = -1, so there are three stationary points at (0,0) , (0,1) , and (0,-1) , correct?

    Next, if:

    <br />
x^2 + y^2 + 1 = 0,<br />

    i.e.

    <br />
y^2 = -x^2 - 1<br />

    then (E.2) becomes... uh...

    <br />
-8y = 0<br />
    this is where I run into trouble.

    Is x \ge 0 ?

    How do I go about finding other points? Are there other stationary points?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276
    Note that for real x and y, x^2\ge0 and y^2\ge0, thus x^2+y^2\ge0, and so x^2+y^2+1\ge1, and thus x^2+y^2+1 is never zero.

    --Kevin C.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,865
    Thanks
    744
    Hello, billym!

    A small error in your derivatives, but your intentions were correct.


    f(x,y) \:=\: x^4 + 2x^2y^2 + y^4 + 4x^2 - 4y^2 + 6
    f_x\;=\;4x^3 + 4xy^2 + {\color{red}8}x \;=\;4x(x^2+y^2+2)

    f_y\;=\;4x^2y + 4y^2 - {\color{red}8}y \;=\;4y(x^2+y^2-2)


    We have: . \begin{array}{cccc}x(x^2 + y^2+2) &=&0 & {\color{blue}[1]} \\<br />
y(x^2+y^2-2) &=&0 & {\color{blue}[2]} \end{array}


    From [1], we have: . \begin{array}{cc}x \:= \:0 \\ x^2+y^2 \:=\:\text{-}2& \leftarrow\text{ impossible}\end{array}

    If x = 0, [2] becomes: . y(y^2-2) \:=\:0\quad\Rightarrow\quad y \:=\:0,\:\pm\sqrt{2}

    . . So far, we have: . (0,\,0),\;\;(0,\,\sqrt{2}),\;\;(0,\,\text{-}\sqrt{2})


    From [2], we have: . \begin{array}{c}y \:=\:0 \\ x^2+y^2 \:=\:2 \end{array}

    If y = 0, [1] becomes: . x(x^2+2) \:=\:0\quad\Rightarrow\quad x \:=\:0

    If x^2+y^2\:=\:2, [1] becomes: . x(2+2) \:=\:0\quad\Rightarrow\quad x \:=\:0
    . . and we've already encountered these solutions.


    Therefore, the stationary points are: . \boxed{(0,\,0),\;\;(0,\,\sqrt{2}),\;\;(0,\,\text{-}\sqrt{2})}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member billym's Avatar
    Joined
    Feb 2008
    Posts
    183

    whoops

    Sorry i made a mistake in the original equation. It actually works out to:

    <br />
f(x,y)= x^4 + 2x^2y^2 + y^4 + 2x^2 - 2y^2 + 6<br />

    So I have a saddle point at (0,0), and local minimums at (0,1) and (0,-1). (I hope).

    Thanks for the help guys.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: August 24th 2011, 12:35 PM
  2. Stationary points with 2 variables
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 3rd 2010, 01:15 PM
  3. Replies: 2
    Last Post: July 21st 2009, 11:49 AM
  4. stationary points of a function of 2 variables
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 10th 2008, 11:38 AM
  5. Replies: 2
    Last Post: February 28th 2008, 06:36 AM

Search Tags


/mathhelpforum @mathhelpforum