Question

The LPP maximize \(z = 2x+5y\) subject to \(x+3y\leq 6\), \(2x+6y\leq 18\), \(x\geq 0\), \(y\geq 0\) has

• A. Unique solution
• B. Infinite solutions
• C. No solution
• D. Unbounded feasible region

Hint:

Draw the feasible region, find its corners and compare z at each corner.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The maximum of a linear objective over a bounded feasible region occurs at a corner point.

Step 2: Simplify the constraints
\(2x+6y\le18\) is the same as \(x+3y\le9\). Since \(x+3y\le6\) is stronger, the second constraint is redundant.

Step 3: Corner points
With \(x,y\ge0\) the region is the triangle with vertices \((0,0)\), \((6,0)\) and \((0,2)\).

Step 4: Evaluate \(z=2x+5y\)
At \((0,0)\): 0. At \((6,0)\): 12. At \((0,2)\): 10.

Step 5: Conclusion
The maximum is 12 at the single point \((6,0)\), so the problem has a unique solution, option (A). The region is bounded, so (D) is wrong, and the objective line is not parallel to an edge, so (B) is wrong.

Final Answer:
The maximum value 12 occurs at only one corner, (6, 0), so the solution is unique, option (A). \[ \boxed{\text{Unique solution}} \]