For the LPP,maximize z=x+4y subject to the co | vedclass

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(D) Given the objective function $z=x+4y$ and the constraints:$1) x+2y \leq 2$$2) x+2y \geq 8$$3) x, y \geq 0$Analyzing the constraints:Constraint $(1

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Has no feasible solution

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(D) Given the objective function $z=x+4y$ and the constraints:$1) x+2y \leq 2$$2) x+2y \geq 8$$3) x, y \geq 0$Analyzing the constraints:Constraint $(1)$ represents the region on or below the line $x+2y=2$,which passes through $(2,0)$ and $(0,1)$. Since $0+2(0) \leq 2$ is true,the region includes the origin.Constraint $(2)$ represents the region on or above the line $x+2y=8$,which passes through $(8,0)$ and $(0,4)$. Since $0+2(0) \geq 8$ is false,the region does not include the origin.Constraint $(3)$ restricts the solution to the first quadrant.Comparing the regions defined by $(1)$ and $(2)$,we observe that the region satisfying $x+2y \leq 2$ and the region satisfying $x+2y \geq 8$ are disjoint. There is no point $(x, y)$ that satisfies both inequalities simultaneously.Therefore,the $LPP$ has no feasible solution.

For the $LPP$,maximize $z=x+4y$ subject to the constraints $x+2y \leq 2$,$x+2y \geq 8$,$x, y \geq 0$.

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