Maximizing $z$ in Equations (1) and (2)

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The forum discussion focuses on maximizing the variable $z$ in the equations $x + y + z = 5$ and $xy + yz + zx = 3$. Participants engaged in solving for the maximum value of $z$ using algebraic manipulation and optimization techniques. The solution provided was confirmed as correct by other users, indicating a consensus on the approach taken. This discussion highlights the importance of understanding constraints in optimization problems.

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$x,y,z\in R$

$x+y+z=5---(1)$

$xy+yz+zx=3---(2)$

find $\max(z)$
 
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Re: find max(z)

My solution:

$\bf{x+y+z=5 \Rightarrow x+y = 5-z...(1)}$$\bf{xy+yz+xz=3}$$\bf{(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)=25 \Rightarrow x^2+y^2+z^2=19}$So $\bf{x^2+y^2 = 19-z^2...(2)}$Using Cauchy - Schwartz Inequality$\bf{\left(x^2+y^2\right).\left(1^2+1^2\right)\geq \left(x+y \right)^2}$$\bf{\left(19 - z^2 \right).2 \geq \left( 5-z \right)^2}$$\bf{3z^2-10z-13 \leq 0}$$\displaystyle \bf{3.\left(z - \frac{13}{3} \right).\left(z+1\right)\leq 0}$$\displaystyle \bf{ -1 \leq z \leq \frac{13}{3}}$

So $\displaystyle \bf{Max.(z) = \frac{13}{3}}$
 
Re: find max(z)

jacks said:
My solution:

$\bf{x+y+z=5 \Rightarrow x+y = 5-z...(1)}$$\bf{xy+yz+xz=3}$$\bf{(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)=25 \Rightarrow x^2+y^2+z^2=19}$So $\bf{x^2+y^2 = 19-z^2...(2)}$Using Cauchy - Schwartz Inequality$\bf{\left(x^2+y^2\right).\left(1^2+1^2\right)\geq \left(x+y \right)^2}$$\bf{\left(19 - z^2 \right).2 \geq \left( 5-z \right)^2}$$\bf{3z^2-10z-13 \leq 0}$$\displaystyle \bf{3.\left(z - \frac{13}{3} \right).\left(z+1\right)\leq 0}$$\displaystyle \bf{ -1 \leq z \leq \frac{13}{3}}$

So $\displaystyle \bf{Max.(z) = \frac{13}{3}}$
thanks , your answer is correct :)
 

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