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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} for $a\in (-5,2)$, we could write it as $\,-5

so

$-5

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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    Alright, thanks for the info! Appreciated!2013-07-24

1 Answers 1

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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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    what about general formula?2012-08-08