The graphical solution set of the system of i | vedclass

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(A) To find the feasible region,we analyze the given system of inequations:$1$. $2x + 3y \leq 6$: The boundary line is $2x + 3y = 6$. For $(0,0)$,$0 \

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Fig. $1$

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(A) To find the feasible region,we analyze the given system of inequations:$1$. $2x + 3y \leq 6$: The boundary line is $2x + 3y = 6$. For $(0,0)$,$0 \leq 6$ is true,so the region is towards the origin.$2$. $x + 4y \geq 4$: The boundary line is $x + 4y = 4$. For $(0,0)$,$0 \geq 4$ is false,so the region is away from the origin.$3$. $x \geq 0, y \geq 0$: This restricts the region to the $1^{	ext{st}}$ quadrant.Combining these,the feasible region is the area bounded by the lines in the $1^{	ext{st}}$ quadrant that satisfies both conditions simultaneously. Looking at the provided figures,Fig. $1$ represents the region bounded by these constraints.Therefore,Option $(A)$ is the correct answer.

The graphical solution set of the system of inequations $2x + 3y \leq 6$,$x + 4y \geq 4$,$x \geq 0$,$y \geq 0$ is given by

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