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

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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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