# Stationary points / two variables (help)

• May 3rd 2008, 04:42 PM
billym
Stationary points / two variables (help)
I have this function and I am looking for the stationary points:

$
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:

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

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

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

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

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

Next, if:

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

i.e.

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

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

$
-8y = 0
$

this is where I run into trouble.

Is $x \ge 0$ ?

How do I go about finding other points? Are there other stationary points?
• May 3rd 2008, 05:25 PM
TwistedOne151
Note that for real x and y, $x^2\ge0$ and $y^2\ge0$, thus $x^2+y^2\ge0$, and so $x^2+y^2+1\ge1$, and thus $x^2+y^2+1$ is never zero.

--Kevin C.
• May 3rd 2008, 05:43 PM
Soroban
Hello, billym!

Quote:

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

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

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

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

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

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

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

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

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

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

• May 3rd 2008, 06:08 PM
billym
whoops
Sorry i made a mistake in the original equation. It actually works out to:

$
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.