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(C) To minimize $z = x + y$ subject to the constraints:$1$) $x + y \geqslant 2$$2$) $x + 2y \leqslant 8$$3$) $y \leqslant 3$$4$) $x, y \geqslant 0$Fir

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infinitely many points

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(C) To minimize $z = x + y$ subject to the constraints:$1$) $x + y \geqslant 2$$2$) $x + 2y \leqslant 8$$3$) $y \leqslant 3$$4$) $x, y \geqslant 0$First,we identify the feasible region by plotting the lines:- $x + y = 2$ passes through $(2, 0)$ and $(0, 2)$.- $x + 2y = 8$ passes through $(8, 0)$ and $(0, 4)$.- $y = 3$ is a horizontal line.The vertices of the feasible region are found by the intersection of these lines:- Intersection of $x + y = 2$ and $x = 0$ is $(0, 2)$.- Intersection of $x + y = 2$ and $y = 0$ is $(2, 0)$.- Intersection of $x + 2y = 8$ and $y = 0$ is $(8, 0)$ (but this is outside the region $y \leqslant 3$ and $x+y \geqslant 2$ constraints).- Intersection of $x + 2y = 8$ and $y = 3$ is $x + 6 = 8 \implies x = 2$,so $(2, 3)$.- Intersection of $x = 0$ and $y = 3$ is $(0, 3)$.The corner points of the feasible region are $(0, 2), (2, 0), (2, 3), (0, 3)$.Evaluating $z = x + y$ at these points:- At $(0, 2)$,$z = 0 + 2 = 2$.- At $(2, 0)$,$z = 2 + 0 = 2$.- At $(2, 3)$,$z = 2 + 3 = 5$.- At $(0, 3)$,$z = 0 + 3 = 3$.The minimum value is $2$,which occurs at all points on the line segment connecting $(0, 2)$ and $(2, 0)$. Since a line segment contains infinitely many points,the solution set contains infinitely many points.

The solution set for minimizing the function $z = x + y$ with constraints $x + y \geqslant 2$,$x + 2y \leqslant 8$,$y \leqslant 3$,$x, y \geqslant 0$ contains

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