Finding extreme values of a variable on an intersection of a sphere and a plane Determine the minimum and maximum value of the variable $z$ defined by the curve given by:
\begin{cases} x^2+y^2+z^2=1 \\ x+2y+2z=0 \end{cases}
So do I need to find a function $z=f(x,y)$ or just find, implicitly, the derivatives that satisfy $f'_x =0, f'_y=0$? I don't know if it is possible to explicitly parametrize the curve with $z=t$, because that is probably how I would usually solve this kind of problem. Any advice?
 A: If Calculus is not mandatory,
Eliminating $x,$
$$y^2+z^2+(-2y-2z)^2=1\iff 5y^2+y(8z)+5z^2-1=0$$
As $y$ is real, the discriminant must be $\ge0$
$\implies(8z)^2\ge20(5z^2-1)\iff z^2\le20/36=5/9$
Now $u^2\le a^2\implies -a\le u\le a$ for $a\ge0$
A: I'll try a different approach without calculus. (Although if it is for your calculus class, you should probably try solving it using methods you have been taught.) 
Substitution
We can rewrite $x^2+y^2+z^2=1$ as
$$\left(\frac{x+2y}{\sqrt5}\right)^2+\left(\frac{y-2x}{\sqrt5}\right)^2+z^2=1.$$
If we denote $u=\frac{x+2y}{\sqrt5}$ and $v=\frac{y-2x}{\sqrt5}$, now we have transformed the problem to maximizing $z$ for
\begin{align}
u^2+v^2+z^2&=1\\
\sqrt5u+2z&=0
\end{align}
So we have $u=-\frac2{\sqrt5}z$ and $u^2+\frac45z^2$. So the first equation is now
$$v^2+\frac95z^2=1.$$
We get maximal possible value for $z$ when $v=0$ and the maximum is $z^2=\frac95$ and $z=\frac3{\sqrt5}$.
Since we know $u$ and $v$, we can also calculate $x$ and $y$ if needed.
Remark. The choice of $u$ and $v$ in the substitution might look somewhat miraculous. However, if you learned something about orthogonal matrices, you should be able to see that this is simply a rotation. We chose such rotation which contains the term $x+2y$, which appears is the second equation.
Geometry
The object we are working with is an intersection of a sphere with radius $1$ and a plane going through the center. So it must be a circle with radius $1$. 
Maybe it would help if we had two dimensional coordinate system in the given plane. The normal vector for this plane is $(1,2,2)$. If we find two perpendicular unit vectors belonging to the plane, they  will give us good coordinate system. For example, vectors $(0,-1,1)$ and $(-4,1,1)$ are perpendicular to each other and also to the normal vector. So they belong to the plane. In order to get unit vectors we divide both vectors by the length and we get the vectors
\begin{align*}
\vec u&=\frac1{\sqrt2}(0,-1,1)\\
\vec v&=\frac1{3\sqrt2}(-4,1,1)
\end{align*}
Then each point of the plane has the form $a\vec u+b\vec v$. The numbers $a$, $b$ are coordinates of this vector in a 2-dimensional cartesian system in the given plane.
If $(x,y,z)=a\vec u+b\vec v$ we get
\begin{align*}
x&=-\frac{4}{3\sqrt2}b\\
y&=-\frac{a}{\sqrt2}+\frac{b}{3\sqrt2}\\
z&=\frac{a}{\sqrt2}+\frac{b}{3\sqrt2}
\end{align*}
The condition $x^2+y^2+z^2=1$ now changes to $a^2+b^2=1$. And we want to maximize $z=\frac{3a+b}{3\sqrt2}$.
Maximizing such expression for points of the circle is a relatively easy problem. Since $z$ increases in the direction of the vector $(3,1)$, we want to "go as far as possible" in this direction. This means that we simply find a vector on the unit circle which is multiple of $(3,1)$. This is the vector $\frac1{\sqrt{10}}(3,1)$, which gives us
\begin{align*}
a&=\frac3{\sqrt{10}}\\
b&=\frac1{\sqrt{10}}\\
z&=\frac{\sqrt{10}}{3\sqrt2}=\frac{\sqrt5}3
\end{align*}
