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.
Best Answer
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,
$$k_{max}=\frac{\sqrt6}{18},\>\>\>\>\>k_{min}=-\frac{\sqrt6}{18}$$