In a Linear Programming Problem (L.P.P.), the | vedclass

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(B) Step $1$: Identify the boundary lines: $L_1: 3x - y = 6$, $L_2: x = 3$, $L_3: y = 2$, $L_4: y = 0$, $L_5: x = 0$.Step $2$: Find intersection point

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$(\frac{8}{3}, 2), (3, 2), (3, 0), (2, 0)$

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(B) Step $1$: Identify the boundary lines: $L_1: 3x - y = 6$, $L_2: x = 3$, $L_3: y = 2$, $L_4: y = 0$, $L_5: x = 0$.Step $2$: Find intersection points of these lines that satisfy all constraints.- Intersection of $L_1$ and $L_2$: $3(3) - y = 6 \implies y = 3$. Since $y \leq 2$, this point $(3, 3)$ is outside.- Intersection of $L_1$ and $L_3$: $3x - 2 = 6 \implies 3x = 8 \implies x = 8/3$. Point is $(8/3, 2)$.- Intersection of $L_1$ and $L_4$: $3x - 0 = 6 \implies x = 2$. Point is $(2, 0)$.- Intersection of $L_2$ and $L_3$: Point is $(3, 2)$.- Intersection of $L_2$ and $L_4$: Point is $(3, 0)$.Step $3$: The feasible region is bounded by the vertices $(2, 0), (3, 0), (3, 2), (8/3, 2)$.

In a Linear Programming Problem ($L$.$P$.$P$.), the corner points of the feasible region determined by the constraints $3x - y \geq 6$, $x \leq 3$, $y \leq 2$, $y \geq 0$, and $x \geq 0$ are:

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