# Thread: [SOLVED] Finding stationary points.

1. ## [SOLVED] Finding stationary points.

so i have a function

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

I want to find the stationary point

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

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

Then to find if its a max or min

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

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

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

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

I want to find the stationary point

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

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

Then to find if its a max or min

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

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

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 $\displaystyle f_x=2x$ and $\displaystyle f_y=3y(2-y)$, we have that the stationary points are $\displaystyle (0,0)$ and $\displaystyle (0,2)$.

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 $\displaystyle M=f_{xx}f_{yy}-f_{xy}^2$

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

Check out the Wikipedia page on the second derivative test.

5. So since M = 12-12y

How do i figure what M equals? Do i use both values for y from each stationary point and put them in equation? So...12-12(0) and 12-12(2)?

6. Yes