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Math Help - Computing critical points - multivariable equation

  1. #1
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    Computing critical points - multivariable equation

    I have the following problem to find critical points for and note whether I have any local minimas, saddle points, or local maximas.

    f(x,y)=x^2y+2y^2-2xy+6

    I found the following partial derivatives:

    f_{x}=2xy-2y and f_{y}=x^2+4y-2x

    Solving 2xy-2y=0, I got: y=0 and x=1

    BUT, I can't figure out all of the points for:
    x^2+4y-2x=0

    I'm getting: y=\frac{1}{2}x-\frac{x^2}{4} ??

    How do you get any critical points from that?


    Any assistance will be appreciated!
    Thank you in advance!
    Jen
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jenberl View Post
    I have the following problem to find critical points for and note whether I have any local minimas, saddle points, or local maximas.

    f(x,y)=x^2y+2y^2-2xy+6

    I found the following partial derivatives:

    f_{x}=2xy-2y and f_{y}=x^2+4y-2x

    Solving 2xy-2y=0, I got: y=0 and x=1

    BUT, I can't figure out all of the points for:
    x^2+4y-2x=0
    plug in y = 0 and solve for x. that gives you one coordinate. then plug in x = 1 and solve for y, that gives you another coordinate.

    to be safe, you can try to find the zeroes of f_y as well, and plug those into f_x. note that f_y is quadratic in x
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    RE: Computing critical points - multivariable equation

    Quote Originally Posted by Jhevon View Post
    plug in y = 0 and solve for x. that gives you one coordinate. then plug in x = 1 and solve for y, that gives you another coordinate.
    Ok.
    So, when I do that for: f_{y}= x^2+4y-2x=0:
    I get x=2 when y=0, and y=\frac{3}{4} when x=2
    But....I don't understand how this works.
    There seems to be an infinite set of answers for this equation which will = 0.
    Does this mean there are no critical points for f_{y}.


    to be safe, you can try to find the zeroes of f_y as well, and plug those into f_x. note that f_y is quadratic in x
    I'm sorry...I am unsure what you mean here. I thought I already computed the critical point for f_{x}: (1,0) Am I wrong?


    Thanks Jhevon, for helping me with this.

    ~Jen
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jenberl View Post
    Ok.
    So, when I do that for: f_{y}= x^2+4y-2x=0:
    I get x=2 when y=0, and y=\frac{3}{4} when x=2
    But....I don't understand how this works.
    There seems to be an infinite set of answers for this equation which will = 0.
    Does this mean there are no critical points for f_{y}.
    say y = 0, then for f_y = 0 we have

    x^2 - 2x = 0 \implies x = 0 \text{ or }x = 2. thus we have two critical points here: (0,0) and (2,0)

    say x = 1, then for f_y = 0 we have

    1 + 4y - 2 = 0 \implies y = \frac 14. thus we have \left( 1, \frac 14 \right) as another critical point.

    now go on to classifying them

    I'm sorry...I am unsure what you mean here. I thought I already computed the critical point for f_{x}: (1,0) Am I wrong?
    forget what i said there, i think i am just overkilling this. my point was that since x^2 + 4y - 2x = 0, we have (by the quadratic formula) x = \frac {2 \pm \sqrt{4 - 16y}}2

    that will probably yield the same solutions we had anyway.
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