<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] As suggested by John Molokach: solving for $z^2$ yields
$$
2x^2+3y^2=30,
$$
so the intersection of both curves projected in the $x,y$ plane is the ellipse 
$$
\begin{cases}
x=\sqrt{15}\cos t\\
y=\sqrt{10}\sin t
\end{cases}
$$
Replacing these equations in the cone yields
$$
z=\sqrt{15+5\sin^2t}
$$
which has maximum $z_{MAX}=2\sqrt{5}$ with elementary derivative analysis.
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