Can anyone find the critical points of this multivariate problem?

Question

f(x,y)= (x-y)/(x^2+2y^2+6)

Answer 1

Critical points are where both partial derivatives df/dx and df/dy equal zero.

df/dy = [ 2y^2 - 4xy - x^2 - 6 ] / [ x^2 + 2y^2 + 6 ]^2

df/dx = [ 2y^2 + 2xy - x^2 + 6 ] / [ x^2 + 2y^2 + 6 ]^2

In order for both partial dervatives to equal zero we need to solve the system

(Eq1) 2y^2 - 4xy - x^2 - 6 = 0
(Eq2) 2y^2 + 2xy - x^2 + 6 = 0.

Adding the 2 equations we get
4y^2 - 2xy - 2x^2 = 0

divide both sides by 2 and factor to get
( 2y + x ) ( y - x ) = 0

Thus we have 2 possibilities to explore.
y = x
x = - 2y

Sunstituting y = x into Eq1 we get
2x^2 - 4x^2 - x^2 - 6 = 0
- 3x^2 - 6 = 0
x^2 = - 2

This has no real solutions.

The other possibility was x = -2y. Substituting that into Eq1 we get

2y^2 + 8y^2 - 4y^2 - 6 = 0
6y^2 - 6 = 0
y = -1, 1

when y = - 1, x = -2(-1) = 2
when y = 1, x = -2(1) = -2

Thus we have 2 critical points

(-2, 1) and (2 , -1)

Good Luck!

Answer 2

fx= 1/(x^2+2y^2+6)^2 * [(x^2+2y2+6)-(x-y)*2x]=0
fy= 1/(x^2+2y^2+6)^2 *[-(x^2+2y^2+6)-((x-y)*4y]=0
x^2+2y^2 +6 -2x*(x-y)=0
-(x^2+2y^2+6)-4y((x-y)=0
sum
(x-y)(x+2y)=0 so x=y and x=-2y
for x=y 3x^2+6= 0 impossible
for x=-2y
6y^2+6+4y(-3y) =0 y^2=1 and y =+-1
the points are (-2,1) and (2,-1)