Q: Given the surface, find a point $(a,b,c)$ with $c>0$, which has a maximum distance from the plane $z=0$. 
Given this surface: $x^2+y^2+z^2+xz-1=0$. 
find a point $(a,b,c)$ with $c>0$, such that it's distance from the plane $z=0$ is maximal.

My attempt:
The Plan: writing the distance from the point $(a,b,c)$ to the plane $z=0$ as a function, and finding the maximum value, and making sure it's on the given surface. 

Distance from a point $(a,b,c)$ to $z=0$ : $|c|=c \Rightarrow$    (since $c>0$)
I'm really not sure how to continue, I don't know what next step I should do, I thought of substituting $c$ in the main function, but, what do I get from that? how to I find $a,b$? I'm basically stuck here. 
Any help is really appreciated, thanks in advance!

A: The surface is given by $x^2+y^2+z^2+xz-1=0$. As you rightly said the distance from a point $(a, b, c)$ on the surface to plane $z = 0$ is $|c|$. So the distance is maximum when $|c|$ is maximum, with the given constraint $a^2 + b^2 + c^2 + ca - 1 = 0$. We only need to consider $c \gt 0$. Applying Lagrange Multiplier method,
$c = \lambda (a^2 + b^2 + c^2 + ca - 1)$ and differentiating we get the following set of equations,
$ (2a + c) \lambda = 0$
$ 2b\lambda = 0$
$ (2c+a)\lambda = 1$
$a^2 + b^2 + c^2 + ca - 1 = 0$
It is clear that $b = 0$. This is something we knew even before we set up these equations.
From first equation, $c = - 2a$ and from third equation, we get $ \displaystyle a = - \frac{1}{3\lambda}, c =  \frac{2}{3\lambda}$
Now if you plug values of $a, b, c$ in the constraint, you get two values of $\lambda$ and you can check which gives maximum $c \ (\gt 0)$.
Can you take it from here?
