If x and y are integers such that (x+1)^2 less than equal to

If x and y are integers such that (x+1)^2 less than equal to 36 and (y-1)^2 less than 64. What is the largest possible and minimum possible value of xy.

\((x+1)^2\leq{36}\) --> \({-\sqrt{36}}\leq{x+1}\leq{\sqrt{36}}\) --> \({-6}\leq{x+1}\leq{6}\) --> \({-7}\leq{x}\leq{5}\).

\((y-1)^2<{64}\) --> \({-\sqrt{64}}<{y-1}<{\sqrt{64}}\) --> \({-8}<{y-1}<{8}\) --> \({-7}<{y}<{9}\), as \(y\) is an integer we can rewrite this inequality as \({-6}\leq{y}\leq{8}\).

We should try extreme values of \(x\) and \(y\) to obtain min and max values of \(xy\):

Min possible value of \(xy\) is for \(x=-7\) and \(y=8\) --> \(xy=-56\);
Max possible value of \(xy\) is for \(x=-7\) and \(y=-6\) --> \(xy=42\).

Solving with absolute values gives the same results:

\((x+1)^2\leq{36}\) means \(|x+1|\leq{6}\) --> \({-7}\leq{x}\leq{5}\).
\((y-1)^2<{64}\) means \(|y-1|<{8}\) --> \({-7}<{y}<{9}\).

Hope it's clear.