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(D) To determine the linear constraints,we analyze the boundaries of the shaded region:$1$. The vertical line passing through $x=1$ with the shaded re

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$x \geqslant 1, y \leqslant 3, x-2y \leqslant 2, 6x+7y \leqslant 42, x \geqslant 0, y \geqslant 0$

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(D) To determine the linear constraints,we analyze the boundaries of the shaded region:$1$. The vertical line passing through $x=1$ with the shaded region to the right gives the constraint $x \geqslant 1$.$2$. The horizontal line passing through $y=3$ with the shaded region below it gives the constraint $y \leqslant 3$.$3$. The line passing through $(2, 0)$ and $(0, -1)$ has the equation $\frac{x}{2} - \frac{y}{1} = 1$,which simplifies to $x - 2y = 2$. Since the shaded region is above this line (e.g.,testing point $(3, 1)$ gives $3 - 2(1) = 1 \leqslant 2$),the constraint is $x - 2y \leqslant 2$.$4$. The line passing through $(7, 0)$ and $(0, 6)$ has the equation $\frac{x}{7} + \frac{y}{6} = 1$,which simplifies to $6x + 7y = 42$. Since the shaded region is below this line (e.g.,testing point $(1, 1)$ gives $6(1) + 7(1) = 13 \leqslant 42$),the constraint is $6x + 7y \leqslant 42$.$5$. The non-negativity constraints are $x \geqslant 0$ and $y \geqslant 0$.Combining these,the correct set of constraints is $x \geqslant 1, y \leqslant 3, x-2y \leqslant 2, 6x+7y \leqslant 42, x \geqslant 0, y \geqslant 0$.

The shaded area in the figure below is the solution set for a certain linear programming problem. Then the linear constraints are given by:

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