Consider the LPP, Max Z = 2x + y, subject to the conditions
\(3x+2y \le 6\)
\(4x+y \le 4\)
\(x \ge 0,\ y \ge 0\), then the maximum value of the objective function is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{16}{5}\)
The feasible region is bounded, so the maximum is at a corner point.
On the x-axis (y = 0): \(3x \le 6\) gives \(x \le 2\), and \(4x \le 4\) gives \(x \le 1\). The tighter one wins, so the corner is (1, 0).
On the y-axis (x = 0): \(2y \le 6\) gives \(y \le 3\), and \(y \le 4\). The corner is (0, 3).
The lines 3x + 2y = 6 and 4x + y = 4 meet where y = 4 − 4x:
\(3x + 2(4 - 4x) = 6\)
\(-5x = -2\)
\( \displaystyle x = \frac{2}{5},\quad y = 4 - \frac{8}{5} = \frac{12}{5} \)
Values of Z = 2x + y at the corners:
\(Z(0, 0) = 0\)
\(Z(1, 0) = 2\)
\(Z(0, 3) = 3\)
\( \displaystyle Z\left(\frac{2}{5}, \frac{12}{5}\right) = \frac{4}{5} + \frac{12}{5} = \frac{16}{5} \)
The largest value is \(\frac{16}{5}\), at \(\left(\frac{2}{5}, \frac{12}{5}\right)\).
Why the other options are incorrect
Option (1) 3 — Incorrect. 3 is the value of Z at the corner (0, 3), which is less than \(\frac{16}{5} = 3.2\).
Option (3) \(\displaystyle\frac{14}{5}\) — Incorrect. It is not the value of Z at any corner; the corner values are 0, 2, 3 and \(\frac{16}{5}\).
Option (4) \(\displaystyle\frac{21}{5}\) — Incorrect. It is larger than every corner value, so Z never reaches it in the feasible region.