For the LP problem,"Maximize z = x + 4y subje | vedclass

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(D) The given constraints are:$1) 3x + 6y \leq 6 \implies x + 2y \leq 2$$2) 4x + 8y \geq 16 \implies x + 2y \geq 4$$3) x \geq 0, y \geq 0$Consider the

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has no feasible region

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(D) The given constraints are:$1) 3x + 6y \leq 6 \implies x + 2y \leq 2$$2) 4x + 8y \geq 16 \implies x + 2y \geq 4$$3) x \geq 0, y \geq 0$Consider the lines $x + 2y = 2$ and $x + 2y = 4$.These are parallel lines. The first inequality $x + 2y \leq 2$ represents the region on or below the line $x + 2y = 2$.The second inequality $x + 2y \geq 4$ represents the region on or above the line $x + 2y = 4$.Since there is no point $(x, y)$ that satisfies both $x + 2y \leq 2$ and $x + 2y \geq 4$ simultaneously,there is no common region that satisfies all the constraints.Therefore,the problem has no feasible region.

For the $LP$ problem,"Maximize $z = x + 4y$ subject to $3x + 6y \leq 6$,$4x + 8y \geq 16$ and $x \geq 0, y \geq 0$."

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