Discussion
The graph given below shows the bounded feasible region (in shaded portion) for the problem :
Max. \(z = 2x_{1} + x_{2} - 12,\)
\(x_{1}, x_{2} \ge 0.\)
The objective function attains its maximum when :
Answer: B
The maximum is attained at \(x_{1} = 3\) and \(x_{2} = 4\).
The constant \(-12\) does not affect where the maximum lies, so compare \(2x_{1} + x_{2}\) at the corners: (0, 0) gives 0, P(4, 0) gives 8, Q(4, 2) gives 10, R(2, 6) gives 10, S(1, 7) gives 9 and T(0, 7) gives 7.
The edge from R to Q has slope \(-2\), the same as the objective's level lines \(2x_{1} + x_{2} = c\), so every point of that edge is optimal with value 10 (z = \(-2\)). The marked point (3, 4) lies on it: \(2 \cdot 3 + 4 = 10\).
(1, 7) gives only 9, (3, 3) is interior and gives 9, and (3, 5) lies outside the region, beyond the edge RQ, so it is not feasible.
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