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Stationary points for surface

  • Thread starter custer
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Find the stationary points for the surface. And find the local and absolute maxima or minima for the following function.
[tex]\f(x,y)= xye^{-1/2(x^2+y^2)}[/tex]

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

and

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

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 [tex]y - x = y^2 - x^2[/tex] from the two equations and that's it.

How do I proceed? Did i do anything wrong?
 

Answers and Replies

  • #2
tiny-tim
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Hi custer! :smile:

(type \partial not \delta for ∂ :wink:)
Find the stationary points for the surface. And find the local and absolute maxima or minima for the following function.
[tex]\f(x,y)= xye^{-1/2(x^2+y^2)}[/tex]

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

and

[tex]\delta{z}/\delta{x}[/tex] = [tex]x[e^{-1/2(x^2+y^2)} - y^2e^{-1/2(x^2+y^2)}][/tex]
You got y = x2 and x = y2, and you combined them by adding (or subtracting?) them …

try substituting one into the other instead. :wink:
 
  • #3
13
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Hi custer! :smile:

(type \partial not \delta for ∂ :wink:)


You got y = x2 and x = y2, and you combined them by adding (or subtracting?) them …

try substituting one into the other instead. :wink:
I think it should be
[tex]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[/tex]
[tex]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[/tex]
 
  • #4
tiny-tim
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oops!

I think it should be
[tex]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[/tex]
[tex]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[/tex]
oops! you're right, I misread it :redface:

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). :smile:
 

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