# Stationary points for surface

1. May 15, 2009

### 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

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

I'm supposed to equate the two equations to get the stationary points (x , y, z) But I couldn't 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?

2. May 15, 2009

### tiny-tim

Hi custer!

(type \partial not \delta for ∂ )
You got y = x2 and x = y2, and you combined them by adding (or subtracting?) them …

try substituting one into the other instead.

3. May 15, 2009

### custer

I think it should be
$$y[e^{-1/2(x^2+y^2)} - x^2e^{-1/2(x^2+y^2)}]= y(1- x^2) e^{-1/2(x^2+ y^2)}= 0$$
$$x[e^{-1/2(x^2+y^2)} - y^2e^{-1/2(x^2+y^2)}]= x(1- y^2)e^{-1/2(x^2+ y^2)}= 0$$

4. May 15, 2009

### tiny-tim

oops!

oops! you're right, I misread it

ok … so you have y(1- x2) = 0 and x(1- y2) = 0 …

solve the first equation (there'll be three solutions), and then substitute (each solution into the second equation).