Stationary Points: Finding Stationary Points of f(x,y)

[gtfitzpatrick]

Homework Statement

determine the stationary points of the function f(x,y) = (x^2 + 1/2(xy) + y^2)e^(x+y)

Homework Equations

The Attempt at a Solution

first i got

df/dx= (x^2+(1/2)xy+y^2)e^(x+y)+(2x+(1/2)y)e^(x+y)

df/dy= (x^2+(1/2)xy+y^2)e^(x+y)+((1/2)x+2y)e^(x+y)

i then let them both = 0

and i get

(x^2+2x+(1/2)xy+(1/2)y+y^2)=0

and

(x^2+(1/2)x+(1/2)xy+2y+y^2)=0

i've tried different to solve them simultaniously and then sub back in. I've tried to factorise them which i think i should do but can't seem to get it could anyone give me some pointers please?

 

[EnumaElish]

Using the quadratic formula you can solve df/dx = 0 as y \in {1/4 (-1 - Sqrt[1 - 32 x - 16 x^2 - 8 xy]), 1/4 (-1 + Sqrt[1 - 32 x - 16 x^2 - 8 xy])}, then sub into df/dy = 0.

 

[HallsofIvy]

Homework Statement

determine the stationary points of the function f(x,y) = (x^2 + 1/2(xy) + y^2)e^(x+y)

Homework Equations

The Attempt at a Solution

first i got

df/dx= (x^2+(1/2)xy+y^2)e^(x+y)+(2x+(1/2)y)e^(x+y)

df/dy= (x^2+(1/2)xy+y^2)e^(x+y)+((1/2)x+2y)e^(x+y)

i then let them both = 0

and i get

(x^2+2x+(1/2)xy+(1/2)y+y^2)=0

and

(x^2+(1/2)x+(1/2)xy+2y+y^2)=0

Subtracting the second from the first gives you (3/2)x- (3/2)y= 0 or y= x. Put that back into either equation (because of the symmetry) gives you x^2+ 2x+ (1/2)x^2+ (1/2)x+ x^2= (5/2)x^2+ (5/2)x= 0 or x(x+ 1)= 0.

i've tried different to solve them simultaniously and then sub back in. I've tried to factorise them which i think i should do but can't seem to get it could anyone give me some pointers please?

 

[HallsofIvy]

Homework Statement

determine the stationary points of the function f(x,y) = (x^2 + 1/2(xy) + y^2)e^(x+y)

Homework Equations

The Attempt at a Solution

first i got

df/dx= (x^2+(1/2)xy+y^2)e^(x+y)+(2x+(1/2)y)e^(x+y)

df/dy= (x^2+(1/2)xy+y^2)e^(x+y)+((1/2)x+2y)e^(x+y)

i then let them both = 0

and i get

(x^2+2x+(1/2)xy+(1/2)y+y^2)=0

and

(x^2+(1/2)x+(1/2)xy+2y+y^2)=0

Subtracting the second from the first gives you (3/2)x- (3/2)y= 0 or y= x. Put that back into either equation (because of the symmetry) gives you x^2+ 2x+ (1/2)x^2+ (1/2)x+ x^2= (5/2)x^2+ (5/2)x= 0 or x(x+ 1)= 0.

i've tried different to solve them simultaniously and then sub back in. I've tried to factorise them which i think i should do but can't seem to get it could anyone give me some pointers please?