Question

For the L.P.P. Maximize \[ z=10x+6y \] subjected to: \[ 3x+y\leq12 \] \[ 2x+5y\leq34 \] \[ x,y\geq0 \] Then the feasible region represented by system of inequalities is:

• A. Unbounded in first quadrant
• B. Bounded in first quadrant
• C. Unbounded in second quadrant
• D. Not possible (Empty)

Hint:

For L.P.P., inequalities with: \[ x\geq0,\ y\geq0 \] usually restrict the feasible region to the first quadrant.

Answer 1

The Correct Option is B

Step 1: Identify the constraints Given: \[ 3x+y\leq12 \] \[ 2x+5y\leq34 \] and: \[ x\geq0,\qquad y\geq0 \] The conditions: \[ x\geq0,\ y\geq0 \] restrict the feasible region to the: \[ \text{first quadrant} \]
Step 2: Analyze the inequalities Both inequalities are of the form: \[ ax+by\leq c \] Hence the feasible region lies: \[ \text{below both lines} \]
Step 3: Check boundedness Since: \[ x\geq0,\qquad y\geq0 \] and both inequalities restrict values of \(x\) and \(y\), the feasible region is enclosed by:
• coordinate axes
• line \(3x+y=12\)
• line \(2x+5y=34\) Hence the feasible region is: \[ \text{bounded} \]
Step 4: Identify the correct option Therefore, the feasible region is: \[ \text{bounded in first quadrant} \] Option analysis:
• Option (A): Incorrect
• Option (B): Correct
• Option (C): Incorrect
• Option (D): Incorrect Hence: \[ \boxed{\text{(B)}} \]