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Math Help - Find maxima, mimima and saddle points (fraction)

  1. #1
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    Find maxima, mimima and saddle points (fraction)

    Find maxima, mimima and saddle points for each of the following func-
    tions:
    a) f(x,y)=6x^2 - 2x^3 + 3y^2 + 6xy
    b) f(x,y)=(x^2 + y^2 - 1)^-1

    Firstly, is my answer for part a) correct:
    df/dx= 12x - 6x^2 + 6y
    df/dy= 6y + 6x

    found x=0 and x=1 then y=0 and y=-1
    so, the critical points are (0,0) and (1.-1)

    doing the 'second derivative test' found (0,0) to be a local minimum and (1,-1) to be a saddle point.

    Secondly, i am at a total loss of what to do for part b) as its a fraction and i cant seem to use partial fractions to seperate the variables!!?

    Thank you x

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  2. #2
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    Quote Originally Posted by cheeseontoast View Post
    Find maxima, mimima and saddle points for each of the following func-

    tions:

    b) f(x,y)=(x^2 + y^2 - 1)^-1


    Secondly, i am at a total loss of what to do for part b) as its a fraction and i cant seem to use partial fractions to seperate the variables!!?

    Thank you x

    If I am not wrong, it is:
    df/dx = -(2x)/(x^2+y^2-1)^2
    df/dy = -(2y)/(x^2+y^2-1)^2 ,isnt it?
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  3. #3
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    Quote Originally Posted by Kiki View Post
    If I am not wrong, it is:
    df/dx = -(2x)/(x^2+y^2-1)^2
    df/dy = -(2y)/(x^2+y^2-1)^2 ,isnt it?
    ok thanks. and what about the rest of the question (finding critical points and maxima/minima)?
    how do i find d2f/dx2 and d2f/dy2?
    Thanks
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