MHB Maximum Value of $x$ with Given Constraints?

[anemone]

Let $x,\,y,\,z$ be real numbers such that $x+2y+3z=86$ and $x^2+y^2+z^2=2014$.

Find the maximum of $x$.

 

[juantheron]

My Solution:

Spoiler

Given $2y+3z = 86-x$ and $y^2+z^2 = 2014-x^2$

Now Using Cauchy-Schwartz Inequality, We Get

$\displaystyle (2^2+3^2)\cdot (y^2+z^2)\geq (2y+3z)^2\Rightarrow 13\cdot (2014-x^2)\geq (86-x)^2$

So $\displaystyle (7x-303)\cdot (x+31)\leq 0$

So $\displaystyle -31\leq x \leq \frac{303}{7}$

 

[anemone]

My Solution:

Spoiler

Given $2y+3z = 86-x$ and $y^2+z^2 = 2014-x^2$

Now Using Cauchy-Schwartz Inequality, We Get

$\displaystyle (2^2+3^2)\cdot (y^2+z^2)\geq (2y+3z)^2\Rightarrow 13\cdot (2014-x^2)\geq (86-x)^2$

So $\displaystyle (7x-303)\cdot (x+31)\leq 0$

So $\displaystyle -31\leq x \leq \frac{303}{7}$

Bravo, jacks!(Yes) And thanks for participating!:)