Consider the LPP: Maximise \(Z = 3x + 2y\) subject to \(x + 2y \le 10\), \(3x + y \le 15\), \(x \ge 0,\ y \ge 0\). Which of the following is true?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → The LPP has infinitely many feasible solutions and exactly one optimal solution.
Every point inside or on the boundary of the feasible region is a feasible solution. The region here is a quadrilateral, so there are infinitely many feasible solutions.
The lines \(x + 2y = 10\) and \(3x + y = 15\) meet where, putting \(y = 15 - 3x\),
\(x + 30 - 6x = 10\)
\(x = 4,\ y = 3\)
The corner points are (0, 0), (5, 0), (4, 3) and (0, 5). Values of Z:
\(Z(0, 0) = 0\)
\(Z(5, 0) = 15\)
\(Z(4, 3) = 12 + 6 = 18\)
\(Z(0, 5) = 10\)
The largest value 18 occurs at one corner only, so the optimal solution (4, 3) is unique.
Why the other options are incorrect
Option (2) The LPP has exactly four feasible solutions and exactly one optimal solution. — Incorrect. Four is the number of corner points; points such as (1, 1) are feasible too.
Option (3) The LPP has infinitely many feasible solutions and infinitely many optimal solutions. — Incorrect. The maximum 18 occurs at (4, 3) only.
Option (4) The LPP has exactly one feasible solution, namely (4, 3). — Incorrect. (4, 3) is the only optimal solution, not the only feasible one.