Relations & Functions
Linear Programming
Grade 12
Question:
<p>The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let <em>z</em> = <em>px</em> + <em>qy</em>. Condition on <em>p</em> and <em>q</em> so that the maximum of <em>z</em> occurs at both the points (15, 15) and (0, 20) is</p>
<p>\(p = q\)</p>
<p>\(p = 2q\)</p>
<p>\(q = 2p\)</p>
<p>\(q = 3p\)</p>
Step-by-Step Solution
Key Concept: For the maximum of a linear objective function z = px + qy to occur at two corner points, the objective function line must be parallel to the edge connecting those two points. This means the slope of the objective line must equal the slope of the edge between (15,15) and (0,20).
<p><strong>Step 1:</strong> For maximum to occur at both (15,15) and (0,20), the line z = px + qy must be parallel to the line segment joining these two points.</p><p><strong>Step 2:</strong> Find the slope of line through (15,15) and (0,20):<br/>Slope = (20-15)/(0-15) = 5/(-15) = -1/3</p><p><strong>Step 3:</strong> The objective function z = px + qy can be written as y = -(p/q)x + z/q<br/>Slope of objective line = -p/q</p><p><strong>Step 4:</strong> For parallelism: -p/q = -1/3<br/>Therefore: p/q = 1/3<br/>Or: <strong>p : q = 1 : 3</strong><br/>Which gives: <strong>3p = q</strong></p><p><strong>Step 5:</strong> Verify: At (15,15): z = 15p + 15q = 15p + 45p = 60p<br/>At (0,20): z = 0 + 20q = 20(3p) = 60p ✓</p><p>∴ Answer: <strong>3p = q</strong> (or equivalently p:q = 1:3)</p>
Correct Answer: C