Lagrange multiplier - space probe i am stuck on this question which uses the Lagrange multiplier. I am trying to construct the equations using the partial derivatives but the $x$'s and $y$'s cancel. can anyone help?
A space probe in the shape of the ellipsoid
$x^2 + y^2 + 3z^2 = 3$
enters a planet's atmosphere and begins to heat up. The temperature on its surface is found
to be
$T(x, y, z) = x^2 + 2y^2 + 6z$:
Use the method of Lagrange multipliers to find the hottest points on the probe's surface.
 A: Set $\Lambda \colon\mathbb R^4\to \mathbb R, (x,y,z,\lambda)\mapsto x^2+2y^2+6z+\lambda (x^2+y^2+3z^2-3)$.
Let $(x,y,z,\lambda)\in \mathbb R^4$.
The following holds:
$$\begin{cases} \Lambda _x(x,y,z,\lambda)&=2x+2\lambda x\\ \Lambda _y(x,y,z,\lambda)&=4y+2\lambda y\\ \Lambda_z(x,y,z, \lambda)&=6+6\lambda _z\\ \Lambda _\lambda(x,y,z,\lambda)&=x^2+y^2+3z^2-3.\end{cases}$$
Suppose $$\begin{cases} 0&=2x+2\lambda x\\ 0&=4y+2\lambda y\\ 0&=6+6\lambda z\\ 0&=x^2+y^2+3z^2-3,\end{cases}$$
then $$\begin{cases} 0&=(1+\lambda)x\\ 0&=(2+\lambda)y\\ \lambda&=-\dfrac1 z \land z\neq 0\neq \lambda \\ 0&=x^2+y^2+3z^2-3,\end{cases}$$
which implies $$\begin{cases} \lambda =-1\lor x=0\\ \lambda =-2 \lor y=0\\ \lambda=-\dfrac1 z \land z\neq 0\neq \lambda \\ x^2+y^2+3z^2-3=0.\end{cases}$$
$\boxed{\text{Case }\lambda =-1}$
It follows that $y=0$ and $z=1$. Thus $x^2+0+3-3=0$ and $x=0$, yielding the critical point $\color{blue}{(0,0,1)}$.
$\boxed{\text{Case }x=0}$


*

*$\boxed{\text{Sub case }\lambda =-2}$ It comes $z=\dfrac 1 2$ and $x^2+y^2-\dfrac 9 4=0$, that is $y=\pm\dfrac 3 2$, yielding the critical points $\color{blue}{\left(0, \dfrac 3 2, \dfrac 1 2\right)}$ and $\color{blue}{\left(0, -\dfrac 3 2, \dfrac 1 2\right)}$.

*$\boxed{\text{Sub case }y=0}$ It comes $0+0+3z~2-3=0$, that is, $z=\pm 1$, yielding the critical points $\color{blue}{(0,0,-1)}$ and $\color{blue}{(0,0,1)}$.


Now just check where it is hotter.
A: Taking the gradient of $T$,
$$\nabla T=(T_x,T_y,T_z)=(2x,4y,6)$$
Taking the gradient of $g$ times $\lambda$,
$$\lambda\nabla g=(\lambda g_x,\lambda g_y,\lambda g_z)=(\lambda 2x,\lambda 2,\lambda g_x)$$
Setting them equal
$$2x=\lambda 2x \Rightarrow 2x(1-\lambda)=0$$
$$4y=\lambda 2y \Rightarrow 2y(2-\lambda)=0$$
$$6=\lambda 6z \Rightarrow \lambda=\frac1{z}$$
So what if $\lambda=1$?  Then from the second equation, $2y=0$ and thus $y=0$.  Also, fron the third equation, we have $1=\frac1{z}$ and thus, $z=1$.  Plugging into the constraint equation, we have 
$$x^2+0^2+3(1)^2=3$$
$$x^2+3=3$$
$$x^2=0$$
$$x=0$$
Thus the critical point when $\lambda =1$ is $(0,0,1)$.
What if $\lambda=2$?  Then from the first equation, $2x(-1)=-2x=0$ and thus $x=0$.  Also, from the third equation, we have $2=\frac1{z}$ and thus $z=\frac1{2}$.  Plugging into the constraint equation we have
$$0^2+y^2+3\left(\frac1{2}\right)^2=3$$
$$y^2+\frac3{4}=3$$
$$y^2=\frac9{4}$$
$$y=\pm\frac3{2}$$
Thus the critical points when $\lambda =2$ is $(0,\pm\frac3{2},\frac1{2})$.
In the case that $x=0$ and $y=0$, we can simply plug $x$ and $y$ into the constraint equation and solve for $z$
$$0^2+0^2+3z^2=3$$
$$3z^2=3$$
$$z^2=1$$
$$z=\pm1$$
Thus the critical points when $x=0,y=0$ is $(0,0,\pm1)$.
Finally, plugging the critical points into $T$
$$T(0,0,1)=0^2+2\cdot0^2+6\cdot 1=6$$
$$T(0,0,-1)=0^2+2\cdot0^2+6\cdot(-1)=-6$$
$$T\left(0,\pm\frac3{2},\frac1{2}\right)=0^2+2\cdot\left(\pm\frac3{2}\right)^2+6\left(\frac1{2}\right)=\frac9{2}+3=\frac{15}{2}$$
