Find the highest point of intersection

Question

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 $(0,\sqrt{10},\sqrt{20})$....?

Answer 1

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 ($0$,$\pm\sqrt{10}$, $\sqrt{20}$)).

Answer 2

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.