How do I find and classify stationary points?

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Homework Help Overview

The discussion revolves around finding and classifying stationary points of a polynomial function involving two variables. The original poster presents their attempts at calculating the partial derivatives and identifying potential stationary points.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to find stationary points by setting partial derivatives to zero and expresses difficulty in the next steps. Participants question the original poster's understanding of the combinations of values for x and y derived from the equations.

Discussion Status

Participants have provided clarifications regarding the combinations of stationary points and the implications of the lack of mixed terms in the polynomial. There is acknowledgment of the identified stationary points, and a question about classification methods has been raised.

Contextual Notes

The original poster indicates uncertainty about the classification of the stationary points and whether additional values are needed for their analysis.

MMCS
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Homework Statement


See attached for problem

Attempt at solution

∂f/∂x = 3x2+12x
∂f/∂y = 3y2-12y

Factorise

x(3x+12)

x=0 or,
3x+12 = 0 → x= -4

y(3y-12)

y=0 or,
3y-12=0 → y=4

I find the next stage difficult, because there are no mixed terms in the polynomial any of the values i substitute into another gives me 0, because of this i need help finding the stationary points, I'm fine with classifying them

Thanks
 

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MMCS said:
because there are no mixed terms in the polynomial any of the values i substitute into another gives me 0, because of this i need help finding the stationary points,

I don't see the difficulty. You have two values for x and two for y. How many combinations does that give for (x, y)?
 
haruspex said:
I don't see the difficulty. You have two values for x and two for y. How many combinations does that give for (x, y)?

This just gives me two stationary points doesn't it?
 
haruspex said:
I don't see the difficulty. You have two values for x and two for y. How many combinations does that give for (x, y)?

I see, it is 4, i thought there would be other values i would need, so my points would be

(0,0)
(0,4)
(-4,0)
(-4,4)

Is this correct?
 
If you had had "mixed" terms, xy, then you would have equations with both x and y leading to the result that a specific value of x leads to a specific value of y. Here, that does not happen. The derivative with respect to x will be 0 if x= 0 or -4 no matter what y is. The derivative with respect to y will be 0 if y= 0 or 4 no matter what x is. In particular, both derivatives will be 0 at (0, 0), (0, 4), (-4, 0), and (-4, 4) as you say.
 
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Thanks, that explanation helps me understand that better
 
MMCS said:
I see, it is 4, i thought there would be other values i would need, so my points would be

(0,0)
(0,4)
(-4,0)
(-4,4)

Is this correct?

Yes, those are the stationary points. As for classification: have you taken (multivariate) second-order tests yet?
 

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