The constraints x+y leq 4,3x+3y geq 18,x geq | vedclass

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(D) To determine the feasible region,we analyze the given constraints:$1$. $x + y \leq 4$$2$. $3x + 3y \geq 18 \implies x + y \geq 6$$3$. $x \geq 0, y

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no feasible region

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(D) To determine the feasible region,we analyze the given constraints:$1$. $x + y \leq 4$$2$. $3x + 3y \geq 18 \implies x + y \geq 6$$3$. $x \geq 0, y \geq 0$From the first constraint,the region must lie on or below the line $x + y = 4$.From the second constraint,the region must lie on or above the line $x + y = 6$.Since these two lines are parallel and represent disjoint half-planes ($x+y \leq 4$ and $x+y \geq 6$),there is no point $(x, y)$ that can satisfy both inequalities simultaneously.Therefore,the feasible region does not exist.

The constraints $x+y \leq 4$,$3x+3y \geq 18$,$x \geq 0$,$y \geq 0$ define:

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