Find the stationary points of $f(x,y,z)= xy+yz+xz$ subject to the constraint $x+y+z-3=0$

Question

So far this is what I have done:

$F(x,y,z)= (xy+yz+xz) + λ(x+y+z-3)$

$Fx = (y+z)+λ=0$

$Fy = (x+z)+λ=0$

$Fz = (y+x)+λ=0$

$Fλ = (x+y+z-3)=0$

I found that $x,y$ and $z$ are $1$ and $λ$ is $-2$

I'm not sure what to do after this.

Answer 1

Let's call the constraint equation $g $, so $g=x+y+z-3$ and we're looking at the level curve $g=0$. By Lagrange multipliers were looking for when the system $$\nabla F=\lambda \nabla g$$ $$g=0$$ Is satisfied. Expanded out that is $$y+z=\lambda$$ $$x+z=\lambda$$ $$y+x=\lambda $$ $$x+y+z-3=0$$ And from there it's just a matter of solving that system for $(x,y,z) $ which satisfies it and those points will be your stationary/critical points.
By simply looking at this system we can gather that $x=y=z$. Using that, the equation $g=0$ tells us that $3x-3=0$ which shows $x=1$, and therefore $(1,1,1) $ satisfies the system and is a critical point.
The work you had done is correct all you have to remember is the solutions to the system of equation you created are the critical points: Just make sure they make sense and you're done.