<s>[INST] Find the highest point of intersection 
Find the highest point of intersection of the sphere $x^2+y^2+z^2=30$ and the cone $x^2+2y^2-z^2=0$.

Am I supposed to use the Lagrange multiplier for this?
EDIT:
So this is what I've tried...
$z^2=-x^2-y^2+30$ and $z^2=x^2+2y^2$. Then $-x^2-y^2+30 = x^2+2y^2$. 
This gives us $3y^2=-2x^2+30$, which is then: $y^2=-\frac23x^2+10$. Substitute $y^2$ in to the sphere equation, we have:
$z^2 = -x^2 - (-\frac23x^2+10)+30 = -\frac13x^2+20$, so the maximum $z$ is $\sqrt{20}$...?
If this is true, then calculating the rest of the point, we have $(0,\sqrt{10},\sqrt{20})$....?
 [/INST] By using Lagrange multipliers, we need to solve system $$F'_x=0,$$ $$F'_y=0,$$ $$F'_z=0,$$ where is $F(x,y,z,\alpha,\beta)=z+\alpha(x^2+y^2+z^2-30)+\beta(x^2+2y^2-z^2)$, 
with conditions $$x^2+y^2+z^2-30=0,$$ $$x^2+2y^2-z^2=0.$$ 
Stationary points are ($\pm \sqrt{15}$,$0$,$\pm \sqrt{15}$) and ($0$,$\pm\sqrt{10}$, $\pm\sqrt{20}$). 
We get that maximal z is $z=\sqrt{20}$ (in point ($0$,$\pm\sqrt{10}$, $\sqrt{20}$)).
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