Question

If $a$ and $b$ satisfy the inequality $-1\le a\le 2$ and $-3\le b\le 2$, find the greatest possible value of $(a+b)(b-a)$.

Answer 1

We need the maximum value of $(a+b)(b-a)$, subject to the condition $-1\le a\le 2$ and $-3\le b\le 2$.

$(a+b)(b-a)$ becomes $b^2-a^2$

So, $b^2$ needs to be maximum and $a^2$ needs to be minimum.

$\therefore b$ would be $-3$ and $a$ would be $0$.

$\therefore (a+b)(b-a)=(0-3)(-3-0)=(-3)(-3)=9$