Trying to find the critical points of $f(x,y) = y^2x - yx^2 + xy$.

I took partial derivative with respect to x, so
$F_x = y^2 - 2xy + y$
$F_x = y(y - 2x + 1)$

Then with respect to y,

$F_y = 2xy - x^2 + x$
$F_y = x( 2y - x + 1 )$

From here I don't know how to find critical points. I've tried solving for $x$ or $y$, but I end up having to cancel out variables which would destroy potential answers. I've also tried setting $F_x = 0$ but four answers gets expanded to 8 answers and I don't know how to go from there. I have a feeling I'm missing something obvious because this is one of the early questions from the book.


Hint Recall that a point $(x_0, y_0)$ is a critical point if $F_x(x_0, y_0) = F_y(x_0, y_0) = 0$. Now, $$F_x(x, y) = y (y - 2 x + 1),$$ and this is zero iff $$y = 0 \qquad \text{or} \qquad y - 2x + 1 = 0.$$ Solving $F_y(x, y) = 0$ gives two similar conditions, giving in all four possible pairs of equations to solve.

  • $\begingroup$ I reached this system of equations but I don't know how to work with them to reduce down to points. I was able to get all the possible answers for X and Y but outside of plugging them in and testing them I didn't know how to reduce them to only correct answers. Was I supposed to just plug them in? $\endgroup$ – Danny Apr 9 '15 at 12:56
  • $\begingroup$ Actually it just clicked, thanks. I actually did it a convoluted way the first time around and it got really messy. $\endgroup$ – Danny Apr 9 '15 at 13:00
  • $\begingroup$ You're welcome, I'm glad you found it useful. $\endgroup$ – Travis Apr 9 '15 at 13:00

You have to set: $$F_x=F_y=0....(1)$$ Then solve the equations and get some $(x,y)$ witch holds $(1)$.


  • $(1,0), (0,1)$ are critical points.

  • $x^2-4xy-x+y^2+y=0$

  • $(x-y)^2-(x+y) -2xy=0$
  • $\begingroup$ How is (0,1) a critical point? $F_x(0,1) = 2$ Also how did you get those 2 equations? Edit: Nevermind, I see how you got them now. $\endgroup$ – Danny Apr 9 '15 at 12:53

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