A system

Algebra Level 4

{ a + b = n 1 a + 1 b = − n 896 m ( n + 896 m ) \large \begin{cases} a+b=n\\ \dfrac{1}{a}+\dfrac{1}{b}=\dfrac{-n}{896m(n+896m)} \end{cases}

Consider the system of equations above. If a − b = 2016 a-b=2016 , what is the maximum possible value of the product m n mn ?


The answer is 567.

Miles Koumouris by Miles Koumouris , 20, Australia

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1 solution

Miles Koumouris
Jan 1, 2017

1 a + 1 b = − n 896 m ( n + 896 m ) \frac{1}{a}+\frac{1}{b}=\frac{-n}{896m(n+896m)}

⇒ n a b = − 4 n 1792 m ( 2 n + 1792 m ) \Rightarrow \frac{n}{ab}=\frac{-4n}{1792m(2n+1792m)}

⇒ − 4 a b = 1792 m ( 2 n + 1792 m ) \Rightarrow -4ab=1792m(2n+1792m)

⇒ n 2 + 1792 m ( 2 n + 1792 m ) = ( a − b ) 2 \Rightarrow n^2 + 1792m(2n+1792m)=(a-b)^2

⇒ ± ( a − b ) = n + 1792 m \Rightarrow \pm(a-b) = n+1792m

⇒ ± 2016 = n + 1792 m \Rightarrow \pm2016 = n+1792m

⇒ m n = m ( ± 2016 − 1792 m ) \Rightarrow mn=m(\pm2016-1792m)

So the maximum value of m n mn is − 1792 × ( 9 16 ) 2 + 2016 × 9 16 = 567 -1792\times (\frac{9}{16})^2 + 2016\times\frac{9}{16} = \boxed{567} .

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