# Thread: [SOLVED] Finding stationary points.

1. ## [SOLVED] Finding stationary points.

so i have a function

$f(x,y) = x^2+3y^2-y^3+5$

I want to find the stationary point

$f_x = 2 , f_y = 6y-3y^2$

so stationary point is at (0,2) ??

Then to find if its a max or min

$f_{xx} = 2 , f_{yy}=6-6y , f_{xy}=0$

So $f_{xx}f_{yy}-f_{xy}^2 = 2 \cdot (6-6y)-(0)^2= 12-12y$

So do i then sub in my value for y e.g y = 2 so ifnd out what that answer is? Or have i done something wrong?

2. Originally Posted by el123
so i have a function

$f(x,y) = x^2+3y^2-y^3+5$

I want to find the stationary point

$f_x = 2 , f_y = 6y-3y^2$

so stationary point is at (0,2) ??

Then to find if its a max or min

$f_{xx} = 2 , f_{yy}=6-6y , f_{xy}=0$

So $f_{xx}f_{yy}-f_{xy}^2 = 2 \cdot (6-6y)-(0)^2= 12-12y$

So do i then sub in my value for y e.g y = 2 so ifnd out what that answer is? Or have i done something wrong?
Given $f_x=2x$ and $f_y=3y(2-y)$, we have that the stationary points are $(0,0)$ and $(0,2)$.

Other than that, what you are doing looks fine.

3. So how do i define if it is a max or min?

4. Originally Posted by el123
So how do i define if it is a max or min?
Define $M=f_{xx}f_{yy}-f_{xy}^2$

If $M>0$ and $f_{xx}>0$, then you have a local minimum.
If $M>0$ and $f_{xx}<0$, then you have a local maximum.
If $M<0$, then you have a saddle point.
If $M=0$, then the test is inconclusive.