Question

In L.P.P., the corner points of the feasible region for the constraints \(3x-y\geq 6,x\leq 3,y\leq 2,y\geq 0,x\geq 0\) are .....

• A. \((3,2),(3,0),(2,0)\)
• B. \((\frac{8}{3},2),(3,2),(3,0),(2,0)\)
• C. \((0,0),(2,0),(\frac{8}{3},2),(0,2)\)
• D. \((3,2),(0,3),(0,2)\)

Hint:

Draw each constraint, find the boundary intersections that lie inside every other constraint, and list them.

Answer 1

The Correct Option is B

Step 1: Test the origin.
At $(0, 0)$: $0 \geq 6$ is false, so the region lies on the far side of the line $3x - y = 6$ from the origin. This removes options with $(0, 0)$, $(0, 2)$ or $(0, 3)$.

Step 2: Check the rest.
Option (A) leaves out the corner where $3x - y = 6$ meets $y = 2$. At $y = 2$: $3x = 8$, $x = 8/3 < 3$, so that corner is in the region.

Step 3: Confirm all four.
$(2, 0)$: $6 \geq 6$ ok. $(3, 0)$: $9 \geq 6$ ok. $(3, 2)$: $7 \geq 6$ ok. $(8/3, 2)$: $6 \geq 6$ ok.

Final Answer:
Option (B). \[ \boxed{\left(\frac{8}{3}, 2\right), (3, 2), (3, 0), (2, 0)} \]