For the linear programming problem: Minimise \(Z = 2x + 4y\) subject to \(x + 2y \ge 10\), \(3x + y \ge 15\), \(x \ge 0\), \(y \ge 0\), the minimum value of Z is attained
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → at infinitely many points
Both constraints are of the “≥” type, so the region is unbounded. The line 3x + y = 15 meets the y-axis at (0, 15), and x + 2y = 10 meets the x-axis at (10, 0). The two lines meet where
\(y = 15 - 3x\)
\(x + 2(15 - 3x) = 10\)
\(30 - 5x = 10\)
\(x = 4,\ y = 3\)
Values of Z at the corner points:
\(Z(0, 15) = 60\)
\(Z(4, 3) = 8 + 12 = 20\)
\(Z(10, 0) = 20\)
For every feasible point,
\(2x + 4y = 2(x + 2y) \ge 20\)
So the minimum is 20, and Z = 20 exactly on the line x + 2y = 10. Every point of the edge joining (4, 3) and (10, 0) gives Z = 20, so the minimum is attained at infinitely many points.
Why the other options are incorrect
Option (1) only at the point (4, 3) — Incorrect. Z is also 20 at (10, 0).
Option (3) only at the point (10, 0) — Incorrect. Z is also 20 at (4, 3), and at points between them such as (6, 2).
Option (4) nowhere, as Z has no minimum value — Incorrect. Z never falls below 20, so the minimum exists although the region is unbounded.