# Thread: Stationary Points - easy or wrong?

1. ## Stationary Points - easy or wrong?

I have to find and classify the stationary points of

$\displaystyle f(x,y)=2x^3-2y^3-3ax^2+3by^2+100$

$\displaystyle a,b \in \Re$

So:

$\displaystyle f_x=6x^2-6ax=0 \Rightarrow 6x^2=6ax \Rightarrow x=a$

$\displaystyle f_y=-6y^2+6by=0+ \Rightarrow-6y^2+6by=0 \Rightarrow y=b$

$\displaystyle f_{xx}=12x-6a$

$\displaystyle f_{yy}=-12y+6b$

$\displaystyle f_{xx}(a,b)=12a-6a=6a$

$\displaystyle f_{yy}(a,b)=-12b+6b=-6b$

$\displaystyle \triangle(a,b)=-36ab$

Have I done this right? How would I classify this?

2. Hello, billym!

I will assume that $\displaystyle a\text{ and }b$ are positive.

Find and classify the stationary points of:
. . $\displaystyle f(x,y)=2x^3-2y^3-3ax^2+3by^2+100 \qquad a,b \in \Re^{{\color{red}+}}$
$\displaystyle f_x \:=\:6x^2 - 6ax \:=\:0 \quad\Rightarrow\quad 6x(x-a) \:=\:0 \quad\Rightarrow\quad x \:=\:0,a$

$\displaystyle f_y \:=\:-6y^2 + 6by \:=\:0 \quad\Rightarrow\quad 6y(b - y) \:=\:0 \quad\Rightarrow\quad y \:=\:0,b$

There are four stationary points to examine: .$\displaystyle (0,0), (a,b),(a,0),(0,b),$

We have: .$\displaystyle \begin{Bmatrix}f_{xx} &=& 12x - 6a \\ f_{yy} &=& -12y + 6b \\ f_{xy} &=& 0 \end{Bmatrix}$

Then: .$\displaystyle D \;=\;6(2x-a)\,(-6)(2y-b) - 0^2 \;=\;-36(2x-a)(2y-b)$

At $\displaystyle (0,0)\!:\;\;D \:=\:(-36)(-a)(-b) \:=\:-36$ . . . saddle point at $\displaystyle (0,0,100)$

At $\displaystyle (a,b)\!:\;\;D \:=\:(-36)(a)(b) \:=\:-36ab$ . . . saddle point at $\displaystyle (a,\:b,\:b^3-a^3+100)$

At $\displaystyle (a,0)\!:\;\;D \:=\:(-36)(a)(-b) \:=\:+36ab$ . . . extreme point
. . At $\displaystyle a,0)\!:\;\;f_{xx} \:=\:12a - 6a \:=\:+6a$ . . . minimum at $\displaystyle (a,\:0.\:100-a^3)$

At $\displaystyle (0,b)\!:\;\;D \:=\:(-36)(-a)(b) \:=\:+36ab$ . . . extreme point
. . At $\displaystyle (0,b)\!:\;\;f_{yy} \:=\:-12b + 6b \:=\:-6b$ . . . maximum at $\displaystyle (0,\:b,\:100 + b^3)$

3. Originally Posted by billym
I have to find and classify the stationary points of

$\displaystyle f(x,y)=2x^3-2y^3-3ax^2+3by^2+100$

$\displaystyle a,b \in \Re$

So:

$\displaystyle f_x=6x^2-6ax=0 \Rightarrow 6x^2=6ax \Rightarrow x=a$

$\displaystyle \color{red}Or\,\,x=0$

$\displaystyle f_y=-6y^2+6by=0+ \Rightarrow-6y^2+6by=0 \Rightarrow y=b$

$\displaystyle \color{red}Or\,\,y=0$

$\displaystyle f_{xx}=12x-6a$

$\displaystyle f_{yy}=-12y+6b$

$\displaystyle f_{xx}(a,b)=12a-6a=6a$

$\displaystyle f_{yy}(a,b)=-12b+6b=-6b$

$\displaystyle \triangle(a,b)=-36ab$

Have I done this right? How would I classify this?
Yes, it looks fiine (just check the cases in red above), and now (a,b) is a saddle point (also called sometimes stationary point) iff $\displaystyle ab>0$, so that the function's Hessian (the matrix whose determinant you denoted by $\displaystyle \Delta$) is indefinite (if you don't know these terms from linear algebra, which would be odd since you're studying multiariable calculus then just forget them...for now).

Tonio