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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


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    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


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    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


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    I got 12-6xy=0 when I added the two equations together. Following the method, I got y^4 + y^2 - 2 = 0.
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