Find the maximum and minimum value of $x$, where:

  • $x+y+z=4$
  • $x^2+y^2+z^2 =6$

I thought I could use these values to form a equation having $x,y,z$ as roots and the sum of roots and $\sum{xy}$ but could not get any idea about the product of roots hence could not move ahead.

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    $\begingroup$ Hint: you are intersecting a plane and a sphere in $\mathbb{R}^3$. The intersection is given by a circle. Can you find its center and radius? $\endgroup$ – Jack D'Aurizio Feb 14 '15 at 10:23
  • $\begingroup$ if you want to brute force it try en.wikipedia.org/wiki/Lagrange_multiplier $\endgroup$ – VigneshM Feb 14 '15 at 10:37

Let we set $x=\frac{4}{3}+u,y=\frac{4}{3}+v,z=\frac{4}{3}+w$. The given constraints translate into: $$\left\{\begin{array}{rcl}v+w &=& -u \\ v^2+w^2 &=& \frac{2}{3}-u^2\end{array}\right.$$ or: $$\left\{\begin{array}{rcl}v+w &=& -u \\ v\cdot w &=& u^2-\frac{1}{3},\end{array}\right.$$ so $v,w$ are roots of the polynomial $p(t)=t^2+ut+\left(u^2-\frac{1}{3}\right)$. They are real roots provided that $u^2-4\left(u^2-\frac{1}{3}\right)\geq 0$, so the critical values for $u$ are given by $u=\pm\frac{2}{3}$ and: $$ x\in\left[\frac{2}{3},2\right].$$

  • $\begingroup$ $p(t=t^2+ut+u^2-1/3$ $\endgroup$ – chenbai Feb 14 '15 at 13:17
  • $\begingroup$ @chenbai: many thanks, now fixed. $\endgroup$ – Jack D'Aurizio Feb 14 '15 at 13:28

Hint for taking off:

Eliminate z between the two equations

$$ x^2 + y^2 -4 x + x y + 4 y + 5 = 0 $$

It is an ellipse. Intersection between a sphere and a plane is a circle, which projects to an ellipse.

Major axis inclined to axes due to presence of $ x y $ term.

Try differentiating with respect to $y$...


My Solution:: Given $$x+y+z=4$$ and $$x^2+y^2+z^2=6$$

So we get $$y+z=(4-x)$$ and $$y^2+z^2 = (6-x^2)$$

Now using the Cauchy-Schwarz inequality, we get

$$\displaystyle \left(y^2+z^2\right)\cdot (1^2+1^2)\geq (y+z)^2$$

So we get $$\displaystyle (6-x^2)\cdot 2 \geq (4-x)^2$$

$$\Rightarrow\displaystyle 12-2x^2\geq 16+x^2-8x$$

$$\Rightarrow\displaystyle 3x^2-8x+4\leq 0\Rightarrow 3x^2-6x-2x+4\leq 0$$

$$\Rightarrow\displaystyle 3x(x-2)-2(x-2)\leq 0\Rightarrow x\in \left[\frac{2}{3}\;,2\right]$$


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