1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Multivariable Functions - stationary points

  1. May 10, 2008 #1
    1. The problem statement, all variables and given/known data



    Find the stationary points of f

    2. Relevant equations

    3. The attempt at a solution

    To find a stationary point the first partial derivative must equal zero, correct?

    I've found the first partial derivatives using the quotient rule but my problem is, I can't rearrange them for y= or x= so I can substitute them into one equation and solve for x or y. What do I do now?
  2. jcsd
  3. May 10, 2008 #2


    User Avatar
    Science Advisor

    Okay, what partial derivatives did you get? What are the equations you are trying to solve?
  4. May 10, 2008 #3
    haha ENG1091? fx=2y^2-2xy+6-x^2 and fy=2y^2+4xy-6+x^2 n then you can put the two equations in quadratic form i.e. fx=(2)y^2-(2x)y+(6-x^2), where a=2, b=-2x and c=6-x^2. Then use the quadratic formula and u should get the value for x and then plug it into the equation u got after using the quadratic formula get y...
    Last edited: May 10, 2008
  5. May 10, 2008 #4
    I have the partial derivatives:



    Now, for a stationary point, both these equations are equal to zero.

    I then have to solve these simultaneously to find values for x, which will in turn give me values for y

    snakeskin - I can see how you've put them in quadratic form and I tried to follow your directions with the quadratic formula, but I got lost. Perhaps i'm doing it wrong.
  6. May 10, 2008 #5
    umm yeah u have to use the quadratic formula for both fx=0 and fy=0 and then u let them equal each other n then u'll have to simplify it which gets a bit tricky. if u use the quadratic formula for fx=0 u should get y=(x +/- (3x^2-12)^0.5)/2 and y=(-2x +/- (6x^2+12)^0.5)/2 for fy=0. When you let (x +/- (3x^2-12)^0.5)/2 = (-2x +/- (6x^2+12)^0.5)/2 you can simplify it to 3x= +/-(6x^2+12)^0.5 -/+(3x^2-12)^0.5. Here I separated the +/- so i get 2 equations to make it easier to handle i.e. 3x= +(6x^2+12)^0.5 -(3x^2-12)^0.5 and 3x= -(6x^2+12)^0.5 +(3x^2-12)^0.5. Now you have to square both sides of the equation and u should get (3x^2-12).(6x^2+12)=0 for both equations n from there u just plug the x value into the equation for y n that should be it... well that's what i did anyways
  7. May 11, 2008 #6


    User Avatar
    Science Advisor

    My first reaction to snakeskin's response was that it couldn't possibly be right- the derivative of a fraction is not a polynomial! But then I realized that he had discarded the (unnecessary) denominator.

    You have
    fx(x,y)=(2y^2-2xy-x^2+6)/(x^2+2y^2+6)^2= 0
    and fy(x,y)=(-2y^2-4xy+x^2+6)/(x^2+2y^2+6)^2= 0 and the first thing you can do is multiply on both sides by the denominators to get 2y^2- 2xy- x^2+ 6= 0 and -2y^1- 4x+ x^2+ 6. Adding those two equations gives -2xy- 4x+ 6= 0 so x(2y+4)= 6 or x(y+ 2)= 3. You could then solve x= 3/(y+ 2) and replace x in either one of the first equations by that. For example, the first equation becomes 2y^2- 6y/(y+ 2)- 9/(y+2)^2+ 6= 0.

    Multiplying by (y+ 2)^2 gives 2y^2(y^2+ 4y+ 4)- 6y(y+2)- 9+ 6(y^2+ 4y+ 4)= 0 which is the same as 2y^4+ 8y^3+ 8y^2- 6y^2- 12y- 9+ 6y^2+ 24y+ 24= 0 or 2y^4+ 8y^3+ 8y^2+ 12y+ 15= 0
  8. May 11, 2008 #7


    User Avatar
    Homework Helper

    I got 12-6xy=0 when I added the two equations together. Following the method, I got y^4 + y^2 - 2 = 0.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook