For the linear programming problem: Maximise \(Z = 6x + 4y\) subject to \(3x + 2y \le 12\), \(x \le 3\), \(x \ge 0\), \(y \ge 0\), the maximum value of Z is attained
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → at infinitely many points
The region is bounded. The line 3x + 2y = 12 meets the y-axis at (0, 6), and meets x = 3 where
\(2y = 12 - 9 = 3\)
\(\displaystyle y = \frac{3}{2}\)
The corner points are (0, 0), (3, 0), \(\left(3, \frac{3}{2}\right)\) and (0, 6).
\(Z(0, 0) = 0\)
\(Z(3, 0) = 18\)
\(\displaystyle Z\left(3, \frac{3}{2}\right) = 18 + 6 = 24\)
\(Z(0, 6) = 24\)
The maximum 24 occurs at two adjacent corners. Since \(6x + 4y = 2(3x + 2y)\), Z equals 24 at every point of the edge \(3x + 2y = 12\) joining them. So the maximum is attained at infinitely many points.
Why the other options are incorrect
Option (1) only at the point (0, 6) — Incorrect. Z is also 24 at \(\left(3, \frac{3}{2}\right)\).
Option (2) at exactly two points — Incorrect. Every point between the two corners on 3x + 2y = 12, such as (2, 3), also gives Z = 24.
Option (4) nowhere, as Z has no maximum value — Incorrect. The region is bounded, so Z has a maximum, namely 24.