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!
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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