Range of variables

Question

Let $x$, $y$ and $z$ be real variables, which satisfy the equation $xy + yz + zx = 7$ and $x+ y+z =6$.

Find range in which the variable lie.

My work : With the given information, I only found out $x^2+y^2+z^2$ and $x^3 + y^3 +z^3 -3xyz$, but this lead to nothing.

What should be my approach?

Answer 1

Let $xyz=a$ and $x\leq y\leq z$.

Thus, $x$, $y$ and $z$ are roots of the equation $f(X)=a$, where $$f(X)=X^3-6X^2+7X.$$ Now, $f'(X)=3X^2-12X+7$, which says that $X_{max}=\frac{6-\sqrt{15}}{3}$, $X_{min}=\frac{6+\sqrt{15}}{3}$ and $$x\leq\frac{6-\sqrt{15}}{3}\leq y\leq \frac{6+\sqrt{15}}{3}\leq z.$$

Now, for $a=f\left(\frac{6-\sqrt{15}}{3}\right)$ we'll get a maximal value of $z$

and for $a=f\left(\frac{6+\sqrt{15}}{3}\right)$ we'll get a minimal value of $x$.

Finally we obtain:

$$2-2\sqrt{\frac{5}{3}}\leq x\leq\frac{6-\sqrt{15}}{3}\leq y\leq \frac{6+\sqrt{15}}{3}\leq z\leq2+2\sqrt{\frac{5}{3}}.$$ Done!

Answer 2

Given $xy+xz+yz=7$ and $x+y+z=6$ find the range of possible values of $x,y,z$.

\begin{eqnarray} (x+y+z)^2&=&36\\ x^2+y^2+z^2+2(xy+xz+yz)&=&36\\ x^2+y^2+z^2+14&=&36\\ x^2+y^2+z^2&=&22\\ x+y+z&=&6 \end{eqnarray}

So we must find the range of values of the coordinates of points lying on the circular intersection of the sphere with center $(0,0,0)$ and radius $\sqrt{22}$ and the plane $x+y+z=6$.

The center of that circular intersection is the point $(2,2,2)$, so we can parameterize the circle as

\begin{equation} (x,y,z)=(2+s,2+t,2-s-t)\tag{$*$} \end{equation}

constrained by the sphere

\begin{eqnarray} (2+s)^2+(2+t)^2+(2-s-t)^2&=&22\\ 2s^2+2st+2t^2+12&=&22\\ s^2+st+t^2&=&5\tag{$**$} \end{eqnarray}

This is an ellipse in the $st$ plane as pictured below.

We see that $s$ achieves maximum and minimum values when $\dfrac{ds}{dt}=0$, that is when

$$ -\frac{s+2t}{2s+t}=0 $$

which is when $s=-2t$.

Substituting this value of $s$ into equation ($**$) gives $t=\pm\sqrt{\dfrac{5}{3}}=\pm\dfrac{\sqrt{15}}{3}$, so $s=\mp\dfrac{2\sqrt{15}}{3}$.

Since $s$ ranges over $\left[-\dfrac{2\sqrt{15}}{3},\dfrac{2\sqrt{15}}{3}\right]$ and since from equation ($**$) we have $x=2+s$ it follows that $x$ ranges over

\begin{equation} \left[2-\frac{2\sqrt{15}}{3},2+\dfrac{2\sqrt{15}}{3}\right]=\left[\frac{6-2\sqrt{15}}{3},\frac{6+2\sqrt{15}}{3}\right] \end{equation}