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.

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.

