Discussion
Consider the Linear Programming Problem (LPP):
Maximize \(\alpha x_{1} + x_{2}\)
Subject to \(2x_{1} + x_{2} \le 6,\)
\(-x_{1} + x_{2} \le 1,\)
\(x_{1} + x_{2} \le 4,\)
\(x_{1} \ge 0, x_{2} \ge 0,\)
where \(\alpha\) is a constant. If \((3, 0)\) is the only optimal solution, then
Answer: D
\((3, 0)\) is the only optimal solution exactly when \(\alpha \gt 2\).
The feasible region has corners \((0, 0)\), \((3, 0)\), \((2, 2)\) (where \(2x_{1} + x_{2} = 6\) meets \(x_{1} + x_{2} = 4\)), \((1.5, 2.5)\) (where \(x_{1} + x_{2} = 4\) meets \(-x_{1} + x_{2} = 1\)) and \((0, 1)\). The objective at each:
\[ \begin{aligned} (3, 0) &: 3\alpha \\ (2, 2) &: 2\alpha + 2 \\ (1.5, 2.5) &: 1.5\alpha + 2.5 \\ (0, 1) &: 1, \quad (0, 0) : 0 \end{aligned} \]
\((3, 0)\) must beat its neighbour \((2, 2)\) strictly: \(3\alpha \gt 2\alpha + 2\), so \(\alpha \gt 2\). Then it also beats the rest. At \(\alpha = 2\) the whole edge from \((3, 0)\) to \((2, 2)\) is optimal, so the optimum is not unique.
For \(1 \lt \alpha \lt 2\) the optimum moves to \((2, 2)\); for \(-2 \lt \alpha \lt 1\) it is at \((1.5, 2.5)\) or \((0, 1)\); and for \(\alpha \lt -2\) the point \((3, 0)\) gives a negative value, far from optimal.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development