For the objective function \(Z = 3x + 9y\) subject to the constraints \(x + 3y \le 60\), \(x + y \ge 10\), \(x \le y\), \(x \ge 0\), \(y \ge 0\), the minimum and the maximum values of Z are
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Minimum 60 and maximum 180
The feasible region lies above \(x + y = 10\), below \(x + 3y = 60\) and above the line \(y = x\). Its corner points:
\(y = x\) and \(x + y = 10\) meet at (5, 5)
\(x + y = 10\) meets the y-axis at (0, 10)
\(x + 3y = 60\) meets the y-axis at (0, 20)
\(y = x\) and \(x + 3y = 60\) meet at (15, 15)
Values of Z:
\(Z(5, 5) = 15 + 45 = 60\)
\(Z(0, 10) = 90\)
\(Z(0, 20) = 180\)
\(Z(15, 15) = 45 + 135 = 180\)
The region is bounded, so the minimum is 60, at (5, 5). The maximum 180 occurs at both (0, 20) and (15, 15), because \(Z = 3(x + 3y)\) is parallel to the edge \(x + 3y = 60\); every point of that edge gives 180.