# Find $a$ for which figure's area is maximum

Suppose that we have following interval $(-5,2)$,we should find such $a$, which takes all possible values from this interval,creates following inequality systems

$$5+a-|2y|\ge 0$$

$$|x|\leq \frac{|a-2|}{2}$$

we are working in $OXY$ cordinantes system,we have to find maximum value of area of figures,which can be defined by all solutions of inequality systems and find possible values of $a$,for which this area is maximum,or shortly we have system of inequality,we have different solution of this system for different value of $a$ from interval $(-5,2)$ and we have different figure created by these different set of solutions,we have to find maximum area between this figures and also this value of $a$ for which this area is possible,i have one idea and dont know if i am correct or wrong,let see first one we have

$$5+a-|2y|\geq 0\Longrightarrow -|2y|\geq -5-a$$ or after dividing by $-2$

$$|y|\leq \frac{5+a}{2}$$

so it means
$$-\frac{5+a}{2}<y<\frac{5+a}{2}$$ for $a\in (-5,2)$, we could write it as $\,-5<y<0\,\,,\,-\frac{6}{2}<y<\frac{6}{2}$

so

$-5<y<0$

for second

$[x]\le[a-2]/2$ $[x]\le7/2$ $[x]\le0$ but last one means that $x=0$ so what i am doing wrong?i think that maximm value could be achieved when $a$ is $-5$ and are is $35/2=17.5$ but i am not sure

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I tried to edit your question but I got tired after a while. Try to learn how to write mathematics with LaTeX in this site ASAP, as otherwise sometimes questions can come out pretty messed up otherwise. –  DonAntonio Aug 7 '12 at 22:51
@Danny Cheuk Currently, my front page has 34 threads which were bumped because you have edited them. It might be a bit too much. For instance, in this case, your edit is very minor and I believe, not crucial. –  1015 Jul 24 '13 at 21:15
Sorry, I thought you can't see what I have edited, I won't do that anymore –  user67258 Jul 24 '13 at 21:21
@DannyCheuk No problem, I tought I'd let you know. Even minor edits bump the question to the front page. I have made a lot of these too until I realized the bumping issue. –  1015 Jul 24 '13 at 21:23
Alright, thanks for the info! Appreciated! –  user67258 Jul 24 '13 at 21:25

Let's take an example, it may clarify things a bit for you. Suppose $a=0$. Then $|y|\le5/2$, and $|x|\le1/2$, so the figure we are talking about is the rectangle bounded by the horizontal lines $y=5/2$ and $y=-5/2$, and the vertical lines $x=-1/2$ and $x=1/2$. This rectangle has sides 5 and 1, and area 5.

Now try it for some other value of $a$, like $a=1$ or $a=-1$, and see what you get.

Then try to get a formula that works for all values of $a$.

EDIT: So, let's finish this one off.

We're told $-5\lt a\lt2$, so $a-2$ is always negative, so $|x|\le(2-a)/2$, so $${a-2\over2}\le x\le{2-a\over2}$$ Also, $5+a$ is always positive, so $|y|\le(5+a)/2$, so $$-{5+a\over2}\le y\le{5+a\over2}$$ So the area in question is a rectangle with sides $2-a$ and $5+a$, hence, area $$(2-a)(5+a)=10-3a-a^2$$ Maximizing the quadratic is an exercise left to the reader.

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does it always form rectangle? –  dato datuashvili Aug 8 '12 at 7:59
Did you try it for some other values of $a$? –  Gerry Myerson Aug 8 '12 at 13:12
sorry no,i was busy –  dato datuashvili Aug 8 '12 at 14:39
what about general formula? –  dato datuashvili Aug 8 '12 at 14:41