## Dhaka Secondary 2011/3

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**don't post problems (by starting a topic)**in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.

### Dhaka Secondary 2011/3

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

### 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.**

### 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

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.### 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

$|x-2| \leq 5$

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.**