Find maxima and minima of $xyz$ given $x+y+z=0$ and $x^2+y^2+z^2=1$ Find the maximum and minimum value of the product of three real numbers $x,y,z$. If the sum of these is equal to zero and the sum of its squares is equal to one.
I supose the following equations:  $$x+y+z=0$$ $$x^2+y^2+z^2=1$$ $$xyz=k$$ 
with $k\in \mathbb R$ the number to find.
So I think that I have to solve the system of equations. The problem is from a chapter of implicit derivates in vector calculus, but I don't even know if I started well. 

By $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$, we find $xy+yz+zx=-\dfrac{1}{2}$.
Now let's take a polinomial function $P(t)=t^3+bt^2+ct+d$ such that real numbers $x,y,z$ are roots of $P$.
By Vieta formulas, we yields $b=0$, $c=-\dfrac{1}{2}$, $d=-k$ and $P(t)=t^3-t\dfrac{1}{2}-k$.
Now let's define $R(t)=t^3-t\dfrac{1}{2}t$. Hence, $R'(t)=3t^2-\dfrac{1}{2}=0$ and $t=\pm\dfrac{1}{\sqrt{6}}$. Therefore local extreme values of $R$: 
$R(\pm\dfrac{1}{\sqrt{6}})=\mp\dfrac{\sqrt{6}}{18}$.
Thus; $x,y,z$ will be real roots of $P$ for $ -\dfrac{\sqrt{6}}{18}\leq k \leq \dfrac{\sqrt{6}}{18}$.
