Find the maximum and minimum value of $f(x,y,z)=x^2+2y^2+3z^2$ in the region $D=\{(x,y,z)\in \mathbb R^3| x^2+y^2+z^2=1\}$. And find a unit vector at which the maximum and minimum are attained respectively.
Attempt:
I know I need to proceed by Lagrange multiplier method, but I am not sure how to proceed after a step
we will get the equations as $$x=\lambda x$$$$y=\frac{\lambda}{2} y$$$$z=\frac{\lambda}{3} z$$$$x^2+y^2+z^2=1$$
Now how to solve them?
Best Answer
This question is a bit tricky given the case analysis you have to do. Always be mindful about dividing by 0.
The first option for $\lambda$ is $\lambda=0$. This gives $x=y=z=0$, which contradicts your constraint equation. So $\lambda\neq 0$. If $x\neq 0$, then $\lambda=1$, giving $y=0,z=0$. With the constraint equation, you get:
$(x,y,z)=(1,0,0)$.
Notice that $y\neq 0$ and $z\neq 0$ is impossible since both would imply $\lambda=0$, and see reasoning above.
There are other cases we've missed, pertaining to other $\lambda$.