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- Feb 14, 2012

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Solve the system of equation

$3x+7y+14z=252\\xyz-u^2=2016$

for non-negative real numbers.

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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,894

-----

Solve the system of equation

$3x+7y+14z=252\\xyz-u^2=2016$

for non-negative real numbers.

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!

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- #2

- Feb 14, 2012

- 3,894

$\begin{align*} 252=3x+7y+14z& \ge 3\sqrt[3]{3x(7y)(14z)}\\&=3\sqrt[3]{3(7)(14)(2016+u^2)}\\& \ge 3\sqrt[3]{3(7)(14)(2016)}\\&=3\sqrt[3]{2^6(3^3)(7^3)}\\&=2^2(3^2)(7)\\&=252\end{align*}$

Equality is attainable when $3x=7y=14z$ and $u=0$. From the first equation of the system, we obtain

$3x=7y=14z=\dfrac{252}{3}=84$. This implies $x=28,\,y=12$ and $z=6$.

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