# Thread: Computing critical points - multivariable equation

1. ## 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.

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

I found the following partial derivatives:

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

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

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

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

How do you get any critical points from that?

Any assistance will be appreciated!
Jen

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

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

I found the following partial derivatives:

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

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

BUT, I can't figure out all of the points for:
$\displaystyle x^2+4y-2x=0$
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 $\displaystyle f_y$ as well, and plug those into $\displaystyle f_x$. note that $\displaystyle f_y$ is quadratic in $\displaystyle x$

3. ## RE: Computing critical points - multivariable equation

Originally Posted by Jhevon
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: $\displaystyle f_{y}= x^2+4y-2x=0$:
I get $\displaystyle x=2$ when $\displaystyle y=0$, and $\displaystyle y=\frac{3}{4}$ when $\displaystyle x=2$
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 $\displaystyle f_{y}$.

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

Thanks Jhevon, for helping me with this.

~Jen

4. Originally Posted by Jenberl
Ok.
So, when I do that for: $\displaystyle f_{y}= x^2+4y-2x=0$:
I get $\displaystyle x=2$ when $\displaystyle y=0$, and $\displaystyle y=\frac{3}{4}$ when $\displaystyle x=2$
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 $\displaystyle f_{y}$.
say y = 0, then for $\displaystyle f_y = 0$ we have

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

say x = 1, then for $\displaystyle f_y = 0$ we have

$\displaystyle 1 + 4y - 2 = 0 \implies y = \frac 14$. thus we have $\displaystyle \left( 1, \frac 14 \right)$ 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 $\displaystyle f_{x}$: (1,0) Am I wrong?
forget what i said there, i think i am just overkilling this. my point was that since $\displaystyle x^2 + 4y - 2x = 0$, we have (by the quadratic formula) $\displaystyle x = \frac {2 \pm \sqrt{4 - 16y}}2$

that will probably yield the same solutions we had anyway.