Stationary Points for Surface: Finding Local and Absolute Extrema

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Homework Help Overview

The discussion revolves around finding stationary points for the function \( f(x,y) = xye^{-1/2(x^2+y^2)} \) and determining local and absolute extrema. Participants are exploring the necessary conditions for stationary points and the implications of their findings.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to derive stationary points by equating partial derivatives but encounters difficulty in progressing beyond a simplified equation. Some participants suggest alternative methods for combining the equations derived from the partial derivatives.

Discussion Status

Participants are actively engaging with the problem, offering suggestions for substitution and clarifying the setup of the equations. There is a recognition of potential misinterpretations in the original equations, and guidance is being provided to explore further solutions.

Contextual Notes

There is an emphasis on correctly interpreting the partial derivatives and ensuring the proper use of notation. The discussion reflects a collaborative effort to clarify the mathematical relationships involved in finding stationary points.

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?
 
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Hi custer! :smile:

(type \partial not \delta for ∂ :wink:)
custer said:
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:
 
tiny-tim said:
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]
 
oops!

custer said:
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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