The maximum value of \(Z\) for the linear programing problem maximize \(Z = x+y\) subject to the constraints \(x+4y \le 8,2x + 3y \le 12, 3x + y \le 9,x \ge 0,y \ge 0\) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(3\displaystyle\frac{10}{11}\)
The maximum of Z on the bounded feasible region occurs at a corner point.
On the x-axis the constraints give \(x \le 8\), \(x \le 6\) and \(x \le 3\), so the corner is (3, 0). On the y-axis they give \(y \le 2\), \(y \le 4\) and \(y \le 9\), so the corner is (0, 2).
The lines \(x + 4y = 8\) and \(3x + y = 9\) meet where
\(y = 9 - 3x\)
\(x + 4(9 - 3x) = 8\)
\(-11x = -28\)
\( \displaystyle x = \frac{28}{11}, \quad y = 9 - \frac{84}{11} = \frac{15}{11} \)
At this point \(2x + 3y = \frac{101}{11} \lt 12\), and the other corners also satisfy it, so \(2x + 3y \le 12\) is a redundant constraint. The corners are (0, 0), (3, 0), \(\left(\frac{28}{11}, \frac{15}{11}\right)\) and (0, 2):
\(Z(0, 0) = 0\)
\(Z(3, 0) = 3\)
\(Z(0, 2) = 2\)
\( \displaystyle Z\left(\frac{28}{11}, \frac{15}{11}\right) = \frac{43}{11} = 3\frac{10}{11} \)
So the maximum value is \(3\frac{10}{11}\).
NCERT Exemplar:
Linear Programming
