Possible solutions: (0,0), (0,6), (4,2)

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Homework Help Overview

The problem involves finding all solutions (x,y) for the equations derived from the partial derivatives of the function f(x,y) = 12xy - x^2y - 2xy^2, specifically where fx(x,y) = 0 and fy(x,y) = 0.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the original poster's attempts to solve the equations derived from the partial derivatives, noting specific solutions found. There is a suggestion to consider (0,0) as a potential solution, prompting further inquiry into how to validate it. One participant points out a possible oversight in the original poster's approach regarding the treatment of y=0.

Discussion Status

The discussion is ongoing, with participants exploring different potential solutions and questioning the completeness of the original poster's findings. There is an acknowledgment of the need to verify the solution (0,0) and a suggestion to re-evaluate the earlier steps taken.

Contextual Notes

Participants are considering the implications of dividing by variables in the equations, which may have led to overlooking certain solutions. There is a focus on ensuring all possible solutions are accounted for without definitive conclusions reached yet.

camino
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Homework Statement



Find all solutions (x,y) for which fx(x,y) = 0 = fy(x,y) if f(x,y) = 12xy - x^2 y - 2xy^2

Homework Equations





The Attempt at a Solution



f(x,y)=12xy-x^2y-2xy^2

fx(x,y)=12y-2xy-2y^2
fy(x,y)=12x-x^2-4xy

0=12y-2xy-2y^2
0=12x-x^2-4xy

EQ 1: 2xy=12y-2y^2
2x=12-2y
x=6-y

EQ 2: 0=12(6-y)-(6-y)^2-4(6-y)y
0=72-12y-(y^2-12y+36)-24y+4y^2
0=3y^2-24y+36
0=3(y^2-8y+12)
0=3(y-6)(y-2)

y=6 y=2
x=0 x=4 so (0,6) , (4,2)

I found those 2 solutions so far, is there any more that I might have missed?
 
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hi camino
i haven't been through your working, but how about (0,0)?
 
I thought about that, but I'm not sure how to prove that as a solution. Any idea?
 
2xy=12y-2y^2
2x=12-2y

You divided by y too early instead of factoring it out. So you lost y=0 as a possible solution. Plug y=0 into both EQ1/2 and see for what x they are both equal to 0.

I think what you did already covers the other solutions that don't have y=0.
 

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