<s>[INST] 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.
 [/INST] 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, 
$$
\left\{
\begin{array}{rlcl}
y\cdot z-\lambda -2x\mu=&0 &\hspace{1cm} & (1)\\
x\cdot z-\lambda -2y\mu=&0  &\hspace{1cm} & (2)\\
x\cdot y-\lambda -2z\mu=&0   &\hspace{1cm} & (3)\\
x+y+z=&10 &\hspace{1cm} & (4)\\
x^2+y^2+z^2=&36  &\hspace{1cm} & (5)\\
\end{array}
\right.
$$
</s>