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(B) $1$. The feasible region is in the first quadrant,so $x \geq 0$ and $y \geq 0$.$2$. The region is bounded on the right by the vertical line passin

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$x, y \geq 0, x-y \geq 0, x \leq 5, y \leq 3$

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(B) $1$. The feasible region is in the first quadrant,so $x \geq 0$ and $y \geq 0$.$2$. The region is bounded on the right by the vertical line passing through $(5, 0)$ and $(5, 3)$,which is $x = 5$. Since the region is to the left of this line,the constraint is $x \leq 5$.$3$. The region is bounded on the top by the horizontal line passing through $(3, 3)$ and $(5, 3)$,which is $y = 3$. Since the region is below this line,the constraint is $y \leq 3$.$4$. The region is bounded on the left by the line passing through the origin $(0, 0)$ and $(3, 3)$. The equation of this line is $y = x$ or $x - y = 0$. Since the shaded region is to the right of this line (e.g.,test point $(4, 1)$ gives $4 - 1 = 3 \geq 0$),the constraint is $x - y \geq 0$.$5$. Combining all these,the constraints are $x, y \geq 0, x - y \geq 0, x \leq 5, y \leq 3$. Thus,option $B$ is correct.

The shaded part of the given figure indicates the feasible region. Then the constraints are:

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