I need your help with the following exercise:
The sphere $x^{2}+y^{2}+z^{2}=4$ is made of certain material whose density is given by $\rho(x,y,z)=y^{2}+xy+2$. I need to find the extreme values of $\rho(x,y,z)$ using the Lagrange multiplier method.
If $g(x,y,z) = x^{2}+y^{2}+z^{2}-4=0$, $\rho_{x}(x,y,z) = y$, $\rho_{y}(x,y,z)=2y+x$, and $\rho_{z}(z,y,z) =0$, then
$$ \begin{align}y &= 2x\lambda \tag{EQ 1} \\ 2y+x &= 2y\lambda \tag{EQ 2} \\ 0 &= 2z\lambda \tag{EQ 3} \\ x^{2}+y^{2}+z^{2}-4&= 0 \tag{EQ 4} \end{align} $$
I'm having trouble solving this system of equations because I'm getting $\lambda=0$ from the third equation.
What does it mean?
Or what am I doing wrong?
OP has derived these : $$ \begin{align}y &= 2x\lambda \tag{EQ 1} \\ 2y+x &= 2y\lambda \tag{EQ 2} \\ 0 &= 2z\lambda \tag{EQ 3} \\ x^{2}+y^{2}+z^{2}-4&= 0 \tag{EQ 4} \end{align} $$
With EQ3 , we get [CASE 1] $\lambda = 0$ or [CASE 2] $z=0$
[CASE 1] then gives $y=0$ , $x=0$ , $z=\pm2$ At these Points , $\rho=0+0+2$
[CASE 2] gives something else.
Plug EQ 1 in EQ 2 to get $2 \times 2x\lambda + x = 2 \times 2x\lambda \times \lambda $
$4 \lambda + 1 = 4 \lambda^2 $
$\lambda = [1 \pm \sqrt{2}]/2$
With that , we then get $x$ & $y$ values (Due to Quadratic Equation & 2 values for $\lambda$ , we get four Solution Points involving $\pm \sqrt{2}$)
We then evaluate $\rho$ , which is $4$ (at all Solution Points) :
[[ generated by Wolfram Online Tool ]]
OVER-VIEW :
[CASE 1] is the Extrema $2$ which occurs at the top & bottom of the Sphere
[CASE 2] is the Extrema $4$ , which occurs Symmetrically on the $XY$ Plane
ERRATA :
The above corresponds to $\rho=x^2+xy+2$ which has a typo.
This corresponds to the correct $\rho=y^2+xy+2$ :
Thus we have multiple relative local Extrema values , while $\rho=6.82843$ is the over-all maxima.
CORRECT OVER-VIEW :
[CASE 1] has the local Extrema $\rho=2$ which occurs at the top & bottom of the Sphere
[CASE 2] has the local Extrema $\rho=1.17157$ & $\rho=6.82843$ , which occurs Symmetrically on the $XY$ Plane