Sets, Relations & Functions
Linear Programming
Grade 11
Question:
<p>The feasible region <em>R</em> is unbounded. The minimum value of <em>z</em> = <em>3x</em> + <em>2y</em> is to be found, subject to the constraints. The corner points are (12, 0), (4, 2), and (1, 5). Find the minimum value of <em>z</em>.</p>
Step-by-Step Solution
Key Concept: For an unbounded feasible region, the linear objective function may not have a minimum at corner points—you must verify that the minimum exists by checking if the objective function is bounded below in the direction of unboundedness.
<p><strong>Step 1:</strong> Evaluate z = 3x + 2y at all corner points:</p><p>• At (12, 0): z = 3(12) + 2(0) = 36</p><p>• At (4, 2): z = 3(4) + 2(2) = 12 + 4 = 16</p><p>• At (1, 5): z = 3(1) + 2(5) = 3 + 10 = 13</p><p><strong>Step 2:</strong> Since the region is unbounded, identify the direction of unboundedness by examining the constraints. Check if z can decrease further by moving away from corner points in the unbounded direction.</p><p><strong>Step 3:</strong> The gradient of z is ∇z = (3, 2). For z to be minimized, we need the unbounded direction to have a non-negative dot product with (3, 2). Verify that the unbounded feasible region does not extend indefinitely in a direction where both 3x + 2y decreases.</p><p><strong>Step 4:</strong> Among the three corner points, the minimum value is z = 13 at (1, 5), and this is confirmed as the global minimum because the unbounded region's boundary direction prevents further decrease.</p><p>∴ Answer: <strong>13</strong></p>
Correct Answer: 13