Computing critical points - multivariable equation

Jenberl · #1 $Old$ March 23rd, 2009, 11:09 PM

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

Jhevon · #2 $Old$ March 23rd, 2009, 11:20 PM

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 $x$

Jenberl · #3 $Old$ March 23rd, 2009, 11:42 PM

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 $f_{y}$ .

Quote:

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

Jhevon · #4 $Old$ March 24th, 2009, 12:01 AM

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

Quote:

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.