Multiplying inequalities of 2 different variable with ranges Given $-\frac{7}{2}\le y <\frac{9}{2}$ and $-1 < x \le 3$, find a and b in $a < xy < b$!
My approach was to find the lowest possible solution for $xy$ which is $3\times-\frac{7}{2}=-\frac{21}{2}$ and to find the highest possible solution for $xy$ which is $3\times\frac{9}{2}$ which is $\frac{27}{2}$. I end up with $-\frac{21}{2} < xy < \frac{27}{2}$.
Is this approach correct?
I also tried multiplying
$0 < x+1 \le 4$ and $0 \le y + \frac{7}{2} < 8$ but this way I get different results. Can I multiply them both if I've made sure that both inequalities are positive, so the signs don't change?
Also, I would love to know if there's another approach I can use in general! Thanks!
 A: I think, your approach is not enough.
Here is one of the possible proofs. 
The right estimation:
Since $|x|\leq3$ and $|y|<\frac{9}{2},$ we obtain: 
$$xy\leq|xy|<\frac{9}{2}\cdot3=\frac{27}{2}.$$
Also, $$xy\geq-\frac{21}{2}.$$
For $x\geq0$ we obtain:
$$xy=x\left(y+\frac{7}{2}\right)-\frac{7}{2}x\geq-\frac{7}{2}\cdot3=-\frac{21}{2}.$$
For $x\leq0$ we obtain:
$$xy=x\left(y-\frac{9}{2}\right)+x\cdot\frac{9}{2}\geq-1\cdot\frac{9}{2}>-\frac{21}{2}.$$
Id est, the best estimation it's:  $$-\frac{21}{2}\leq xy<\frac{27}{2}$$
A: The extrema can occur inside the rectangular domain, on a side or at a corner.
For the inside you need to consider the stationary points, by cancelling the gradient. This gives the value $0$ at the point $(0,0)$.
On the sides, you have non-constant linear functions, and the extrema can only occur at the endpoints, which are the corners.
Finally, it suffices to compare the values at the corners, which confirms
$$-\frac{21}2\le xy<\frac{27}2.$$
Note that these bounds are a minimum and a supremum. We processed the question as if all inequalities were non-strict, and in the end checked whether the value could be reached or not.
