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(C) $1$. Identify the feasible region defined by the constraints: $x + y \geqslant 2$,$x + 2y \leqslant 8$,$y \leqslant 3$,$x, y \geqslant 0$. $2$. Th

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at infinite number of points on the line segment joining $(0, 2)$ and $(2, 0)$

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(C) $1$. Identify the feasible region defined by the constraints: $x + y \geqslant 2$,$x + 2y \leqslant 8$,$y \leqslant 3$,$x, y \geqslant 0$. $2$. The corner points of the feasible region are found by solving the intersection of the lines: - Intersection of $x + y = 2$ and $x = 0$ gives $(0, 2)$. - Intersection of $x + y = 2$ and $y = 0$ gives $(2, 0)$. - Intersection of $x + 2y = 8$ and $y = 3$ gives $(2, 3)$. - Intersection of $x + 2y = 8$ and $x = 0$ gives $(0, 4)$,but $y \leqslant 3$ restricts this to $(0, 3)$. $3$. Evaluate $z = x + y$ at the corner points: - At $(0, 2)$,$z = 0 + 2 = 2$. - At $(2, 0)$,$z = 2 + 0 = 2$. - At $(2, 3)$,$z = 2 + 3 = 5$. - At $(0, 3)$,$z = 0 + 3 = 3$. $4$. The minimum value of $z$ is $2$,which occurs at both $(0, 2)$ and $(2, 0)$. $5$. Since the objective function $z = x + y$ is parallel to the constraint $x + y = 2$,the minimum value occurs at all points on the line segment joining $(0, 2)$ and $(2, 0)$.

The solution for minimizing the function $z = x + y$ under an $L$.$P$.$P$. with constraints $x + y \geqslant 2$,$x + 2y \leqslant 8$,$y \leqslant 3$,$x, y \geqslant 0$ is

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