The region formed by the inequalities x, y ge | vedclass

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(C) The given inequalities are $x \geq 0, y \geq 0, y \leq 6$,and $x+y \leq 3$.$1$. The conditions $x \geq 0$ and $y \geq 0$ restrict the region to th

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bounded in first quadrant

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(C) The given inequalities are $x \geq 0, y \geq 0, y \leq 6$,and $x+y \leq 3$.$1$. The conditions $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.$2$. The line $x+y = 3$ passes through $(3, 0)$ and $(0, 3)$. Since $(0, 0)$ satisfies $x+y \leq 3$,the region lies towards the origin.$3$. The line $y = 6$ is a horizontal line. Since $x+y \leq 3$ already implies $y \leq 3$ (given $x \geq 0$),the condition $y \leq 6$ is redundant as the region is already bounded by $x+y \leq 3$ and the axes.$4$. The intersection of these regions forms a triangle with vertices $(0, 0), (3, 0)$,and $(0, 3)$.Since this region has finite boundaries,it is a bounded region in the first quadrant.

The region formed by the inequalities $x, y \geq 0, y \leq 6, x+y \leq 3$ is $.....$

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