Discussion
Consider the following LPP
Max : x1 + 5/2 x2
Subject to
5x1 + 3x2 = 15;
−x1 + x2 ≤ 1;
2x1 + 5x2 ≤ 10;
x1, x2 ≥ 0
The problem :
Answer: C
The problem has a unique optimal solution.
On the equality \(5x_{1} + 3x_{2} = 15\), write \(x_{1} = 3 - 0.6x_{2}\). The objective becomes
\[ x_{1} + 2.5x_{2} = 3 + 1.9x_{2} \]
which increases with \(x_{2}\). The constraints limit \(x_{2}\): \(-x_{1} + x_{2} \le 1\) gives \(x_{2} \le 2.5\), and \(2x_{1} + 5x_{2} \le 10\) gives \(3.8x_{2} \le 4\), so \(x_{2} \le \frac{20}{19}\). The optimum is the single point \(x_{2} = \frac{20}{19}\), \(x_{1} = \frac{45}{19}\), with value 5.
The trap is option B: the objective is parallel to \(2x_{1} + 5x_{2} \le 10\), which would give a whole edge of optima if the feasible region were two-dimensional. The equality constraint cuts the region down to a segment that meets that line at one point only. The region is non-empty and bounded, so A and D are wrong.
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