MHB Find Maximum $(xyz)^2$ Given $x+y+z=0, x^2+y^2+z^2=2015$

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To find the maximum of $(xyz)^2$ given the constraints $x+y+z=0$ and $x^2+y^2+z^2=2015$, it is essential to express $z$ in terms of $x$ and $y$. Substituting $z = -x - y$ into the second equation leads to a relationship between $x$ and $y$. The maximum value can be derived using methods such as Lagrange multipliers or by substituting specific values that satisfy the conditions. The final result indicates that the maximum of $(xyz)^2$ under these constraints is achievable and can be calculated explicitly. The discussion emphasizes the importance of algebraic manipulation in optimizing the expression.
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If $x,\,y,\,z$ are three real numbers such that $x+y+z=0$ and $x^2+y^2+z^2=2015$, find the maximum of $(xyz)^2$.
 
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anemone said:
If $x,\,y,\,z$ are three real numbers such that $x+y+z=0---(1)$ and $x^2+y^2+z^2=2015---(2)$, find the maximum of $(xyz)^2$.
let $xyz=k$
from $(1)(2):xy+yz+zx=\dfrac {-2015}{2}$
$x,y,z$ are three roots of:$t^3-\dfrac{2015 t}{2}-k=0$
$k=t^3-\dfrac {2015t}{2}$
$max(k)$ occurs when $t^2=\dfrac{2015}{6}$
$\therefore k=t(t^2-\dfrac {2015}{2})=t(\dfrac {-2015}{3})$
and $max(xyz)^2=max(k^2)=\dfrac{2015^3}{54}$
 
Thanks for participating, Albert!:)

Your answer is correct, of course!
 
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