Linear Programming
Feasible Region
Grade 12
Question:
<p>In the shaded region bounded by <em>y</em> = 3<em>x</em> and <em>y</em> = <em>x</em>/2, if <em>P</em>(<em>a</em><sub>1</sub>, <em>a</em><sub>2</sub>) lies in the shaded region, then for any point <em>R</em>(<em>x</em>, <em>y</em>) in the shaded region, \(y > \dfrac{x}{2}\) and \(y < 3x\). If \(P(a, a^2)\) lies in the shaded region, then \(a \in\)?</p>
<p>\(\left(0, 3\right)\)</p>
<p>\(\left(\dfrac{1}{2}, 3\right)\)</p>
<p>\(\left(0, \dfrac{1}{2}\right)\)</p>
<p>\(\left(\dfrac{1}{2}, \infty\right)\)</p>
Step-by-Step Solution
Key Concept: In a linear programming region bounded by two lines through the origin, any point must satisfy BOTH inequalities simultaneously: it must lie above the lower boundary AND below the upper boundary. The feasible region is the intersection of these half-planes, not their union.
<p><strong>Step 1:</strong> Identify the two boundary lines: y = 3x (steeper line) and y = x/2 (gentler line). Both pass through the origin.</p><p><strong>Step 2:</strong> The shaded region bounded by these lines is the region between them, where points satisfy both: y ≥ x/2 (on or above the lower line) AND y ≤ 3x (on or below the upper line).</p><p><strong>Step 3:</strong> For any point P(a₁, a₂) in the shaded region: a₂ ≥ a₁/2 and a₂ ≤ 3a₁ must both hold.</p><p><strong>Step 4:</strong> For ANY other point R(x, y) in this same shaded region, the same constraints apply: y ≥ x/2 AND y ≤ 3x (with equality possible on boundaries).</p><p><strong>Step 5:</strong> The statement 'y > x/2 AND y < 3x' correctly describes the interior of the feasible region (strict inequalities exclude the boundary).</p><p>∴ Answer: B</p>
Correct Answer: B