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.
 A: Consider the equation: $A^3-10A^2+32A-xyz=0$, whose roots are $x,y,z$. We need to find the maximum and minimum values of:
$$xyz=A^3-10A^2+32A=f(A).$$ 
FOC:
$$f'(A)=3A^2-20A+32=0 \Rightarrow A_1=\frac83; \ \ A_2=4.$$
SOC:
$$\begin{align}f''(A)&=6A-20; \\
f''\left(\frac83\right)&=-4<0 \Rightarrow f\left(\frac83\right)=\frac{896}{27} \ (\text{max});\\
f''(4)&=4>0 \Rightarrow f(4)=32 \ (\text{min}).\end{align}
$$
A: We have $$x^2+y^2=36-z^2$$ and $$x+y=10-z,$$ which gives
$$(10-z)^2-2xy=36-z^2$$ or
$$xy=32-10z+z^2$$ and $$xyz=32z-10z^2+z^3.$$
Also, $$(x+y)^2\geq4xy,$$ which gives
$$3z^2-20z+28\leq0$$ or
$$2\leq z\leq\frac{14}{3}.$$
Can you end it now?
I got $\max(xyz)=\frac{896}{27}$ and $\min(xyz)=32$.
A: 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, 
$$
\left\{
\begin{array}{rlcl}
y\cdot z-\lambda -2x\mu=&0 &\hspace{1cm} & (1)\\
x\cdot z-\lambda -2y\mu=&0  &\hspace{1cm} & (2)\\
x\cdot y-\lambda -2z\mu=&0   &\hspace{1cm} & (3)\\
x+y+z=&10 &\hspace{1cm} & (4)\\
x^2+y^2+z^2=&36  &\hspace{1cm} & (5)\\
\end{array}
\right.
$$
A: As an alternative, we can also try to solve the problem by a geometrical intepretation, indeed the constraints represent a circle which is the intersection between


*

*the plane $x+y+z=10$

*the sphere $x^2+y^2+z^2=36$


therefore the center of the circle is located at
$$OC=\left(\frac{10}{3},\frac{10}{3},\frac{10}{3}\right)$$
and the radius of the circle is
$$R^2=36-|OC|^2 \implies R^2=\frac83 \implies R=\frac{2\sqrt 6}{3}$$
To parametrize the circle let consider at first the circle in $x-y$ plane and centered in the origin
$$x^2+y^2=\frac83 \iff \left(\frac{2\sqrt 6}{3}\cos \alpha, \frac{2\sqrt 6}{3} \sin \alpha,0\right)$$
and since the rotation matrix around $u=\left(\frac{\sqrt 2}{2},-\frac{\sqrt 2}{2},0\right)$ of an angle $\theta=\arccos \left(\frac{\sqrt 3}{3}\right)$ is given by
$$M = \begin{bmatrix} 
\frac12+\frac{\sqrt 3}{6} & -\frac12 +\frac{\sqrt 3}{6} & \frac{\sqrt 3}{3} \\ 
-\frac12 +\frac{\sqrt 3}{6} & \frac12+\frac{\sqrt 3}{6} & \frac{\sqrt 3}{3} \\ 
-\frac{\sqrt 3}{3} & -\frac{\sqrt 3}{3} & \frac{\sqrt 3}{3}
\end{bmatrix}$$
the intersection circle can be parametrized as follow


*

*$x(\alpha)=\frac{10}{3}+\frac{2\sqrt 6}{3}\left(\frac12+\frac{\sqrt 3}{6}\right)\cos \alpha  +\frac{2\sqrt 6}{3}\left(-\frac12 +\frac{\sqrt 3}{6}\right)\sin \alpha 
=\frac{\sqrt 6}{3}\left[\frac{5\sqrt 6}3+\left(1+\frac{\sqrt 3}{3}\right)\cos \alpha  +\left(-1+\frac{\sqrt 3}{3}\right)\sin \alpha\right]$

*$y(\alpha)=\frac{10}{3}+\frac{2\sqrt 6}{3}\left(-\frac12 +\frac{\sqrt 6}{6} \right)\cos \alpha +\frac{2\sqrt 6}{3}\left(\frac12+\frac{\sqrt 6}{6}\right)\sin \alpha
=\frac{\sqrt 6}{3}\left[\frac{5\sqrt 6}3+\left(-1+\frac{\sqrt 3}{3}\right)\cos \alpha  +\left(1+\frac{\sqrt 3}{3}\right)\sin \alpha\right]$

*$z(\alpha)=\frac{10}{3}-\frac{2\sqrt 2}{3}\left(\cos \alpha +\sin \alpha \right)=\frac{2\sqrt 2}{3}\left(\frac{5\sqrt 2}{2}-\cos \alpha -\sin \alpha \right)$
and finally we obtain
$$xyz(\alpha)=\frac{8}{27}\left(\sqrt 2 \cos (3\alpha )-\sqrt 2 \sin (3\alpha)+110\right) \quad \alpha \in [0,2\pi)$$
that is

$$xyz(\alpha)=\frac{8}{27}\left[ 2 \sin \left(\frac{\pi}4-3\alpha\right)+110\right] \quad \alpha \in [0,2\pi)$$

and therefore

$$32\le xyz \le \frac{896}{27}$$

form the plot of we can see that the maximum and the minimum values are reached each one in threee different points of the domain.

A: It seems like a better start if you eliminate a variable (or two...)
Thus $z= 10-x-y$, so $x^2+y^2+100-2(x+y)+(x+y)^2=36$, that is, $x^2+y^2+xy-x-y+64=0$.
You are looking for the maximum of $xy(10-x-y)$. 
If you do not want to look for anything creative, Lagrange multiplier surely works. 
