Maximum Value of $x$ with Given Constraints?

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The discussion focuses on maximizing the value of the variable $x$ under the constraints $x + 2y + 3z = 86$ and $x^2 + y^2 + z^2 = 2014$. Participants explored various mathematical approaches to derive the maximum value of $x$, confirming that the solution involves applying techniques from linear algebra and optimization. The consensus indicates that the maximum value of $x$ can be determined using methods such as the Cauchy-Schwarz inequality and Lagrange multipliers.

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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 $x$.
 
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My Solution:
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 $\displaystyle -31\leq x \leq \frac{303}{7}$
 
Last edited by a moderator:
jacks said:
My Solution:
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!:)
 

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