## Dhaka Secondary 2011/3

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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### Dhaka Secondary 2011/3

$|x – 2| \leq 5$ and $y \leq |x – 4|$. Find the minimum and maximum values of $xy$.

Hasib
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### Re: Dhaka Secondary 2011/3

i think the maximum is positive infinity and minimum is negative infinity.
A man is not finished when he's defeated, he's finished when he quits.

Moon
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### Re: Dhaka Secondary 2011/3

Why? We want some logic. You don't need to write formal solution, but at least show some logic.
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

Hasib
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### Re: Dhaka Secondary 2011/3

Ok!
$|x-2| \leq 5$
or, $-3 \leq x \leq 7$
now $y \leq |x-4|$
now, just plug x=-3
we have now $y \leq 7$ so, we can plug y as negative infinity, so xy=positive infinity.

Now, plug x=7 we now have $y \leq 3$ now again plug y as negative infinity. We have xy=negative infinity
A man is not finished when he's defeated, he's finished when he quits.