What is the easiest and simplest way to bound $xy$

Question

Let $-1\le x\le 2$ and $-2\le y\le 3$. I have to bound $xy$. I can do it case by case. E.g

Case1 \begin{cases} 0\le -x\le 1\\ 0\le -y\le2 \end{cases} Taking the product term by term yields $\boxed{0\le xy\le 2}$

Case2 \begin{cases} 0\le -x\le 1\\ 0\le y\le3 \end{cases} Again I take the product to have $\boxed{-3\le xy\le0}$

Case3

\begin{cases} 0\le x\le 2\\ 0\le -y\le2 \end{cases} then $\boxed{-4\le xy\le 0}$

Case 4

\begin{cases} 0\le x\le 2\\ 0\le y\le3 \end{cases} then $\boxed{0\le xy\le6}$

Combining all these stuffs gives finally $-4\le xy\le 6$.

Does someone have an alternative method, more simpler and not as long as this thing above.

Answer 1

Proof without words:

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Answer 2

In this case, it is easy to see that the least value of the product $xy$ is either $-1×3$ or $-2×2.$ Similarly, the greatest value of this product in the rectangle is either $-1×-2$ or $2×3,$ so that we have $$-4\le xy\le 6.$$

Answer 3

$f(x)=x$ has a global negative minimum on interval of definition and $f(y)=y$ a global positive maximum on interval of definition and $f(x)=x$ has a global positive maximum on interval of definition and $f(y)=y$ a global negative minimum on interval of definition and both $f(x)=x$ and $f(y)=y$ are strictly monotone.

Global minimum of $xy$ on that rectangle is then lesser of the product of those two pairs of values.

Similarly for global maximum of $xy$ on that rectangle.