# Finding partial derivatives

1. Mar 19, 2009

### camino

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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?

2. Mar 19, 2009

### lanedance

hi camino

3. Mar 19, 2009

### camino

I thought about that, but i'm not sure how to prove that as a solution. Any idea?

4. Mar 19, 2009

### bob1182006

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.