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}
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