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. 
 A: With $z=-(x+y)$, rewrite the equation $x^2+y^2+z^2=1$ as $(x+y)^2-xy=\frac12$. Then,
$$k=xyz=-xy(x+y)=\frac12(x+y)-(x+y)^3$$
To find the extrema of $k$, take its derivative with respect to $x+y$ and set it to zero, which leads to $x+y=\pm\frac 1{\sqrt6}$. Thus, the maximum and minimum values are, respectively,
$$k_{max}=\frac{\sqrt6}{18},\>\>\>\>\>k_{min}=-\frac{\sqrt6}{18}$$
A: The intersection between the plane $x+y+z=0$ and the sphere $x^2+y^2+z^2=1$ is a circle, which can be parameterized as,
$$(x,y,z)= \frac2{\sqrt6}\left(\cos (t+\frac\pi3),\> \cos (t-\frac\pi3), \>- \cos t\right)$$
which starts at $(\frac1{\sqrt6},\frac1{\sqrt6},-\frac2{\sqrt6})$ and circles counter-clockwise over $0\le t <2\pi$. The product is then,
$$xyz=-\frac4{3\sqrt6}\cos (t+\frac\pi3)\cos (t-\frac\pi3)\cos t = -\frac{\cos 3t}{3\sqrt6}$$
which has the maximum value $\frac1{3\sqrt6}$ when $\cos3t=-1$ and minimum value $-\frac1{3\sqrt6}$ when $\cos3t=1$. Note that the extrema are reached at multiple points along the circle, where their locations can be derived by solving $\cos 3t=\pm1$.
A: Try to interpret x, y, and z as coordinates in a standard three dimensional coordinate system - the first equation defines a plane, the second a sphere. You can then imagine where the sphere and the plane intersect, and how large k would be for all those points.
A: 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}$.
