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Math Help - Stationary Points - easy or wrong?

  1. #1
    Member billym's Avatar
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    Stationary Points - easy or wrong?

    I have to find and classify the stationary points of

    f(x,y)=2x^3-2y^3-3ax^2+3by^2+100

    a,b \in \Re

    So:

    f_x=6x^2-6ax=0 \Rightarrow 6x^2=6ax \Rightarrow x=a

    f_y=-6y^2+6by=0+ \Rightarrow-6y^2+6by=0 \Rightarrow y=b

    f_{xx}=12x-6a

    f_{yy}=-12y+6b

    f_{xx}(a,b)=12a-6a=6a

    f_{yy}(a,b)=-12b+6b=-6b

    \triangle(a,b)=-36ab

    Have I done this right? How would I classify this?
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  2. #2
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    Hello, billym!

    I will assume that a\text{ and }b are positive.


    Find and classify the stationary points of:
    . . f(x,y)=2x^3-2y^3-3ax^2+3by^2+100 \qquad a,b \in \Re^{{\color{red}+}}
    f_x \:=\:6x^2 - 6ax \:=\:0 \quad\Rightarrow\quad 6x(x-a) \:=\:0 \quad\Rightarrow\quad x \:=\:0,a

    f_y \:=\:-6y^2 + 6by \:=\:0 \quad\Rightarrow\quad 6y(b - y) \:=\:0 \quad\Rightarrow\quad y \:=\:0,b


    There are four stationary points to examine: . (0,0), (a,b),(a,0),(0,b),


    We have: . \begin{Bmatrix}f_{xx} &=& 12x - 6a \\ f_{yy} &=& -12y + 6b \\ f_{xy} &=& 0 \end{Bmatrix}


    Then: . D \;=\;6(2x-a)\,(-6)(2y-b) - 0^2 \;=\;-36(2x-a)(2y-b)


    At (0,0)\!:\;\;D \:=\:(-36)(-a)(-b) \:=\:-36 . . . saddle point at (0,0,100)


    At (a,b)\!:\;\;D \:=\:(-36)(a)(b) \:=\:-36ab . . . saddle point at (a,\:b,\:b^3-a^3+100)


    At (a,0)\!:\;\;D \:=\:(-36)(a)(-b) \:=\:+36ab . . . extreme point
    . . At a,0)\!:\;\;f_{xx} \:=\:12a - 6a \:=\:+6a . . . minimum at (a,\:0.\:100-a^3)


    At (0,b)\!:\;\;D \:=\:(-36)(-a)(b) \:=\:+36ab . . . extreme point
    . . At (0,b)\!:\;\;f_{yy} \:=\:-12b + 6b \:=\:-6b . . . maximum at (0,\:b,\:100 + b^3)

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  3. #3
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    Quote Originally Posted by billym View Post
    I have to find and classify the stationary points of

    f(x,y)=2x^3-2y^3-3ax^2+3by^2+100

    a,b \in \Re

    So:

    f_x=6x^2-6ax=0 \Rightarrow 6x^2=6ax \Rightarrow x=a

    \color{red}Or\,\,x=0

    f_y=-6y^2+6by=0+ \Rightarrow-6y^2+6by=0 \Rightarrow y=b

    \color{red}Or\,\,y=0

    f_{xx}=12x-6a

    f_{yy}=-12y+6b

    f_{xx}(a,b)=12a-6a=6a

    f_{yy}(a,b)=-12b+6b=-6b

    \triangle(a,b)=-36ab

    Have I done this right? How would I classify this?
    Yes, it looks fiine (just check the cases in red above), and now (a,b) is a saddle point (also called sometimes stationary point) iff ab>0, so that the function's Hessian (the matrix whose determinant you denoted by \Delta) is indefinite (if you don't know these terms from linear algebra, which would be odd since you're studying multiariable calculus then just forget them...for now).

    Tonio
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