Discussion
Consider the following LPP :
Max. : \(Z = 2x_{1} + x_{2}\)
Subject to :
\(3x_{1} + 4x_{2} \le 6\)
\(6x_{1} + x_{2} \le 3\)
\(x_{1} \ge 0\), \(x_{2} \ge 0\)
What is the solution of this LPP ?
Answer: D
The optimal solution is (2/7, 9/7).
The feasible region has corners (0, 0), (1/2, 0) on \(6x_{1} + x_{2} = 3\), (0, 3/2) on \(3x_{1} + 4x_{2} = 6\), and the intersection of the two lines. From the second, \(x_{2} = 3 - 6x_{1}\); substituting, \(3x_{1} + 12 - 24x_{1} = 6\), so \(x_{1} = 2/7\) and \(x_{2} = 9/7\).
\[ \begin{aligned} Z(1/2, 0) &= 1, \quad Z(0, 3/2) = 1.5 \\ Z(2/7, 9/7) &= 4/7 + 9/7 = 13/7 \end{aligned} \]
The maximum, \(13/7 \approx 1.86\), is at (2/7, 9/7). The other points fail: (5/7, 5/7) violates \(6x_{1} + x_{2} \le 3\) (it gives 35/7), (3/11, 14/11) is feasible but gives only \(Z = 20/11 \approx 1.82\), and (1/3, 2/3) gives 4/3.
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