For the objective function \(Z = 15x + 30y\) subject to the constraints \(x + y \le 8\), \(2x + y \ge 8\), \(x - 2y \ge 0\), \(x \ge 0\), \(y \ge 0\), the maximum and the minimum values of Z are
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Maximum 160 and minimum 60
The feasible region lies below \(x + y = 8\), above \(2x + y = 8\) and below the line \(x = 2y\). Its corner points:
\(2x + y = 8\) meets the x-axis at (4, 0)
\(x + y = 8\) meets the x-axis at (8, 0)
\(x = 2y\) and \(x + y = 8\) meet where \(3y = 8\):
\(\left(\displaystyle\frac{16}{3}, \frac{8}{3}\right)\)
\(x = 2y\) and \(2x + y = 8\) meet where \(5y = 8\):
\(\left(\displaystyle\frac{16}{5}, \frac{8}{5}\right)\)
Values of Z:
\(Z(4, 0) = 60\)
\(Z(8, 0) = 120\)
\(Z\left(\displaystyle\frac{16}{3}, \frac{8}{3}\right) = 80 + 80 = 160\)
\(Z\left(\displaystyle\frac{16}{5}, \frac{8}{5}\right) = 48 + 48 = 96\)
The region is bounded, so the maximum is 160 and the minimum is 60.