Sets, Relations & Functions
Linear Programming
Grade 11
Question:
<p>The objective function \(z\) occurs maximum at (15, 15) and (0, 20). If \(z = 115p + 15q\), find \(q\) in terms of \(p\) such that \(z\) is maximized subject to the conditions at both corner points being equal.</p>
<p>\(q = p\)</p>
<p>\(q = 2p\)</p>
<p>\(q = 3p\)</p>
<p>\(q = 4p\)</p>
Step-by-Step Solution
Key Concept: If the objective function achieves its maximum at two different corner points, those points must lie on the same level curve (isoquant line). This means the objective function coefficients must satisfy a proportionality condition with the coordinates of both points.
<p><strong>Step 1:</strong> If z achieves maximum at both (15, 15) and (0, 20), then z must have the same value at both points.</p><p><strong>Step 2:</strong> Calculate z at (15, 15):<br/>z₁ = 115p(15) + 15q(15) = 1725p + 225q</p><p><strong>Step 3:</strong> Calculate z at (0, 20):<br/>z₂ = 115p(0) + 15q(20) = 0 + 300q = 300q</p><p><strong>Step 4:</strong> Set z₁ = z₂ for maximum to occur at both points:<br/>1725p + 225q = 300q<br/>1725p = 75q<br/>q = 23p</p><p>∴ Answer: C (q = 23p)</p>
Correct Answer: C