The maximum value of \(Z = x + y\) subject to the constraints \(2x + y \le 8\), \(x + 2y \le 8\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{16}{3}\)
The region is bounded, so the maximum of Z is at a corner point. On the axes, 2x + y = 8 gives (4, 0) and x + 2y = 8 gives (0, 4).
The two lines meet where
\(2x + y = x + 2y\), so \(x = y\)
\(3x = 8\)
\(\displaystyle x = y = \frac{8}{3}\)
The corner points are (0, 0), (4, 0), \(\left(\frac{8}{3}, \frac{8}{3}\right)\) and (0, 4).
\(Z(0, 0) = 0\)
\(Z(4, 0) = 4\)
\(\displaystyle Z\left(\frac{8}{3}, \frac{8}{3}\right) = \frac{16}{3}\)
\(Z(0, 4) = 4\)
The maximum is \(\frac{16}{3}\), at \(\left(\frac{8}{3}, \frac{8}{3}\right)\).
Why the other options are incorrect
Option (1) 4 — Incorrect. This is Z at the corners (4, 0) and (0, 4); the corner where the lines meet gives more.
Option (3) 8 — Incorrect. This is Z at (4, 4), which is not feasible: there 2x + y = 12 > 8.
Option (4) 5 — Incorrect. This is Z at the feasible point (2, 3), which is not a corner; \(\frac{16}{3} \approx 5.33\) is larger.