The maximum value of \(Z = 3x + 4y\) subject to the constraints \(x + y \le 9\), \(2x + y \le 14\), \(x + 3y \le 21\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 33
On the x-axis the binding line is \(2x + y = 14\), giving (7, 0). On the y-axis it is \(x + 3y = 21\), giving (0, 7).
\(x + y = 9\) meets \(2x + y = 14\) where
\(x = 5, \; y = 4\)
\(x + y = 9\) meets \(x + 3y = 21\) where
\(2y = 12\)
\(y = 6, \; x = 3\)
Both points satisfy the third constraint: \(5 + 12 = 17 \le 21\) and \(6 + 6 = 12 \le 14\). So the bounded region has the corner points (0, 0), (7, 0), (5, 4), (3, 6) and (0, 7). Values of Z:
\(Z(0, 0) = 0\)
\(Z(7, 0) = 21\)
\(Z(5, 4) = 15 + 16 = 31\)
\(Z(3, 6) = 9 + 24 = 33\)
\(Z(0, 7) = 28\)
The maximum value is 33, at (3, 6).