# Thread: Stationary points / two variables (help)

1. ## Stationary points / two variables (help)

I have this function and I am looking for the stationary points:

$\displaystyle f(x,y) = x^4 + 2x^2y^2 + y^4 + 4x^2 - 4y^2 + 6$

I started by trying to solve the two partial derivatives:

$\displaystyle 4x(x^2 + y^2 + 1) = 0, (E.1)$
$\displaystyle 4y(y^2 + x^2 - 1) = 0, (E.2)$

If $\displaystyle x = 0$, then $\displaystyle (E.2)$ becomes:

$\displaystyle 4y(y^2 - 1) = 0$

Thus $\displaystyle y = 0$ , $\displaystyle y = 1$ , or $\displaystyle y = -1$, so there are three stationary points at $\displaystyle (0,0)$ , $\displaystyle (0,1)$ , and $\displaystyle (0,-1)$ , correct?

Next, if:

$\displaystyle x^2 + y^2 + 1 = 0,$

i.e.

$\displaystyle y^2 = -x^2 - 1$

then $\displaystyle (E.2)$ becomes... uh...

$\displaystyle -8y = 0$
this is where I run into trouble.

Is $\displaystyle x \ge 0$ ?

How do I go about finding other points? Are there other stationary points?

2. Note that for real x and y, $\displaystyle x^2\ge0$ and $\displaystyle y^2\ge0$, thus $\displaystyle x^2+y^2\ge0$, and so $\displaystyle x^2+y^2+1\ge1$, and thus $\displaystyle x^2+y^2+1$ is never zero.

--Kevin C.

3. Hello, billym!

$\displaystyle f(x,y) \:=\: x^4 + 2x^2y^2 + y^4 + 4x^2 - 4y^2 + 6$
$\displaystyle f_x\;=\;4x^3 + 4xy^2 + {\color{red}8}x \;=\;4x(x^2+y^2+2)$

$\displaystyle f_y\;=\;4x^2y + 4y^2 - {\color{red}8}y \;=\;4y(x^2+y^2-2)$

We have: .$\displaystyle \begin{array}{cccc}x(x^2 + y^2+2) &=&0 & {\color{blue}[1]} \\ y(x^2+y^2-2) &=&0 & {\color{blue}[2]} \end{array}$

From [1], we have: .$\displaystyle \begin{array}{cc}x \:= \:0 \\ x^2+y^2 \:=\:\text{-}2& \leftarrow\text{ impossible}\end{array}$

If $\displaystyle x = 0$, [2] becomes: .$\displaystyle y(y^2-2) \:=\:0\quad\Rightarrow\quad y \:=\:0,\:\pm\sqrt{2}$

. . So far, we have: .$\displaystyle (0,\,0),\;\;(0,\,\sqrt{2}),\;\;(0,\,\text{-}\sqrt{2})$

From [2], we have: .$\displaystyle \begin{array}{c}y \:=\:0 \\ x^2+y^2 \:=\:2 \end{array}$

If $\displaystyle y = 0$, [1] becomes: .$\displaystyle x(x^2+2) \:=\:0\quad\Rightarrow\quad x \:=\:0$

If $\displaystyle x^2+y^2\:=\:2$, [1] becomes: .$\displaystyle x(2+2) \:=\:0\quad\Rightarrow\quad x \:=\:0$
. . and we've already encountered these solutions.

Therefore, the stationary points are: .$\displaystyle \boxed{(0,\,0),\;\;(0,\,\sqrt{2}),\;\;(0,\,\text{-}\sqrt{2})}$

4. ## whoops

Sorry i made a mistake in the original equation. It actually works out to:

$\displaystyle f(x,y)= x^4 + 2x^2y^2 + y^4 + 2x^2 - 2y^2 + 6$

So I have a saddle point at (0,0), and local minimums at (0,1) and (0,-1). (I hope).

Thanks for the help guys.