Stationary points for surface

• May 15th 2009, 02:01 AM
custer
Stationary points for surface
Find the stationary points for the surface. And find the local and absolute maxima or minima for the following function.
$f(x,y)= xye^{-1/2(x^2+y^2)}$

Is there any stationary points for this surface?
Let z = f(x,y)
I found the
$\delta{z}/\delta{x}$ = $y[e^{-1/2(x^2+y^2)} - x^2e^{-1/2(x^2+y^2)}]$

and

$\delta{z}/\delta{x}$ = $x[e^{-1/2(x^2+y^2)} - y^2e^{-1/2(x^2+y^2)}]$

I'm supposed to equate the two equations to get the stationary points (X , Y) and substitute them into f(x,y) to get Z. But I could not since I could only get $y - x = y^2 - x^2$ from the two equations and that's it.

How do I proceed? Did i do anything wrong?
• May 15th 2009, 02:11 AM
Spec
$\left\{\begin{array}{ll}f'_x=y(1-x^2)e^{-\frac{1}{2}(x^2+y^2)}=0\\f'_y=x(1-y^2)e^{-\frac{1}{2}(x^2+y^2)}=0\end{array}\right.$

$e^{-\frac{1}{2}(x^2+y^2)} > 0$, so you're left with:

$\left\{\begin{array}{ll}f'_x=y(1-x^2)=0\\f'_y=x(1-y^2)=0\end{array}\right.$
• May 15th 2009, 03:10 AM
custer
Quote:

Originally Posted by Spec
$\left\{\begin{array}{ll}f'_x=y(1-x^2)e^{-\frac{1}{2}(x^2+y^2)}=0\\f'_y=x(1-y^2)e^{-\frac{1}{2}(x^2+y^2)}=0\end{array}\right.$

$e^{-\frac{1}{2}(x^2+y^2)} > 0$, so you're left with:

$\left\{\begin{array}{ll}f'_x=y(1-x^2)=0\\f'_y=x(1-y^2)=0\end{array}\right.$

Sorry. I don't get what you mean. I thought $e^{-\frac{1}{2}(x^2+y^2)}$ will only result between 0 and 1, including 0 and 1.
$e^{-\infty}$ = 0 and $e^0$ = 1

btw, i think i'm confused
• May 15th 2009, 03:12 AM
HallsofIvy
Quote:

Originally Posted by custer
Find the stationary points for the surface. And find the local and absolute maxima or minima for the following function.
$f(x,y)= xye^{-1/2(x^2+y^2)}$

Is there any stationary points for this surface?
Let z = f(x,y)
I found the
$\delta{z}/\delta{x}$ = $y[e^{-1/2(x^2+y^2)} - x^2e^{-1/2(x^2+y^2)}]$

and

$\delta{z}/\delta{x}$ = $x[e^{-1/2(x^2+y^2)} - y^2e^{-1/2(x^2+y^2)}]$

Do you mean $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$?

Quote:

I'm supposed to equate the two equations to get the stationary points (X , Y) and substitute them into f(x,y) to get Z. But I could not since I could only get $y - x = y^2 - x^2$ from the two equations and that's it.
No, you do not "equate the two equations", you set each expression equal to 0.

Quote:

How do I proceed? Did i do anything wrong?
$y[e^{-1/2(x^2+y^2)} - x^2e^{-1/2(x^2+y^2)}]= y(1- x^2) e^{-1/2(x^2+ y^2)}= 0$
$x[e^{-1/2(x^2+y^2)} - y^2e^{-1/2(x^2+y^2)}]= x(1- y^2)e^{-1/2(x^2+ y^2)}= 0$

Since an exponential is never 0, those are easy to solve.

I find 5 stationary points.
• May 15th 2009, 03:16 AM
custer
Quote:

Originally Posted by HallsofIvy
Do you mean $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$?

No, you do not "equate the two equations", you set each expression equal to 0.

$y[e^{-1/2(x^2+y^2)} - x^2e^{-1/2(x^2+y^2)}]= y(1- x^2) e^{-1/2(x^2+ y^2)}= 0$
$x[e^{-1/2(x^2+y^2)} - y^2e^{-1/2(x^2+y^2)}]= x(1- y^2)e^{-1/2(x^2+ y^2)}= 0$

Since an exponential is never 0, those are easy to solve.

I find 5 stationary points.

By equating them I mean they both equals to zero.
Anyway, million thanks :)