Question

The correct constraints for the given feasible region are ...

• A. \(x+2y-2\geq 0,3x+5y-15\geq 0,2x-3y+6\geq 0,x\geq 0,y\geq 0\)
• B. \(x+2y-2\leq 0,3x+5y-15\leq 0,2x-3y+6\leq 0,x\geq 0,y\geq 0\)
• C. \(x+2y-2\leq 0,3x+5y-15\geq 0,2x-3y+6\geq 0,x\geq 0,y\geq 0\)
• D. \(x+2y-2\geq 0,3x+5y-15\leq 0,2x-3y+6\geq 0,x\geq 0,y\geq 0\)

Hint:

Find each line's equation from its intercepts and test the origin to choose the inequality direction.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
From the figure, \(L_{1}\) passes through \((0,1)\) and \((2,0)\), \(L_{2}\) through \((0,3)\) and \((5,0)\), and \(L_{3}\) through \((-3,0)\) and \((0,2)\). The shaded region lies in the first quadrant.

Step 2: Line Equations
\(L_{1}:x+2y=2\), \(L_{2}:3x+5y=15\), \(L_{3}:\dfrac{x}{-3}+\dfrac{y}{2}=1\Rightarrow2x-3y+6=0\).

Step 3: Choose the Side
The shaded region is above \(L_{1}\) (away from the origin), so \(x+2y-2\geq0\). It is below \(L_{2}\) (origin side), so \(3x+5y-15\leq0\). It is below \(L_{3}\) (origin side); at the origin \(2x-3y+6=6\gt0\), so \(2x-3y+6\geq0\).

Step 4: Result
Add \(x\geq0\), \(y\geq0\). This matches option (D).

Final Answer:
The constraints are x + 2y - 2 >= 0, 3x + 5y - 15 <= 0, 2x - 3y + 6 >= 0 with x, y >= 0. \[ \boxed{\text{(D) }x+2y-2\geq 0,\ 3x+5y-15\leq 0,\ 2x-3y+6\geq 0,\ x\geq 0,\ y\geq 0} \]