<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] 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.

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