1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

$\textbf{HINT:}$

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

  • $x^2-4xy-x+y^2+y=0$

  • $(x-y)^2-(x+y) -2xy=0$
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.