Find the smallest and highest value of the product $xyz$ Find the smallest and highest value of the product $xyz$ assuming that:
$x + y + z = 10$ and
$x^2 + y^2 + z^2 = 36$.
I calculated this:
$x+y+z=10 => (x+y+z)^2=10^2$
$x^2+y^2+z^2+2xy+2yz+2zx=100$
$(x^2+y^2+z^2+2xy+2yz+2zx)-(x^2+y^2+z^2)=100-36$
$2xy+2yz+2zx=64$
$xy+yz+zx=32$
I'm stuck. What is the next step to this exercise?
My idea is to show the equation using one variable and after computing the derivative reach global extremes.

Use the the method of Lagrange multipliers. The problem is 
$$
\left\{
\begin{array}{rl}
\min & x\cdot y\cdot z \\
\mbox{such that} & x+y+z=10\\
                 & x^2 + y^2 + z^2 = 36 
\end{array}
\right.
$$ 
By Lagrange multiplier theorem the critical points of $f (x, y, z)=x\cdot y\cdot z$ satisfying the constraints $g(x,y,z)= x+y+z-10=0$  and $h(x,y,z)=x^2+y^2+z^2-36=0$ is solution of 
$$
\left\{
\begin{array}{rl}
\frac{\partial}{\partial x}\mathcal{L}(x,y,z,\lambda,\mu)=&0\\
\frac{\partial}{\partial y}\mathcal{L}(x,y,z,\lambda,\mu)=&0\\
\frac{\partial}{\partial z}\mathcal{L}(x,y,z,\lambda,\mu)=&0\\
\frac{\partial}{\partial \lambda}\mathcal{L}(x,y,z,\lambda,\mu)=&0\\
\frac{\partial}{\partial \mu}\mathcal{L}(x,y,z,\lambda,\mu)=&0\\
\end{array}
\right.
$$
Here, 
$
\mathcal{L}(x,y,z,\lambda,\mu)= f(x,y,z)-\lambda g(x,y,z)-\mu h(x,y,z)
$
In this case, 
$$
\left\{
\begin{array}{rlcl}
y\cdot z-\lambda -2x\mu=&0 &\hspace{1cm} & (1)\\
x\cdot z-\lambda -2y\mu=&0  &\hspace{1cm} & (2)\\
x\cdot y-\lambda -2z\mu=&0   &\hspace{1cm} & (3)\\
x+y+z=&10 &\hspace{1cm} & (4)\\
x^2+y^2+z^2=&36  &\hspace{1cm} & (5)\\
\end{array}
\right.
$$
