The solution set of the constraints |x-y| leq | vedclass

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(B) The given constraints are $|x-y| \leqslant 1$ and $x, y \geqslant 0$. The inequality $|x-y| \leqslant 1$ is equivalent to $-1 \leqslant x-y \leqsl

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an unbounded set

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(B) The given constraints are $|x-y| \leqslant 1$ and $x, y \geqslant 0$. The inequality $|x-y| \leqslant 1$ is equivalent to $-1 \leqslant x-y \leqslant 1$,which can be split into two inequalities: $y \leqslant x+1$ and $y \geqslant x-1$. Combining these with $x \geqslant 0$ and $y \geqslant 0$,we observe the region in the first quadrant. For any $x \geqslant 0$,we can choose $y$ such that $x-1 \leqslant y \leqslant x+1$. As $x$ increases indefinitely,$y$ can also increase indefinitely (e.g.,by choosing $y=x$). Since the region extends infinitely in the first quadrant,the solution set is an unbounded set.

The solution set of the constraints $|x-y| \leqslant 1, x, y \geqslant 0$ is

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