Applications of Derivatives
Linear Programming Problem
Grade 12
Question:
<p>The feasible solution for a LPP is shown in the following figure. Let \(z = 3x - 4y\) be the objective function. Minimum of \(z\) occurs at</p><p>Corner points: \((0,0),\, (0,8),\, (4,10),\, (6,8),\, (6,5),\, (5,0)\)</p>
<p>\((0, 0)\)</p>
<p>\((0, 8)\)</p>
<p>\((5, 0)\)</p>
<p>\((4, 10)\)</p>
Step-by-Step Solution
Key Concept: For a linear objective function in LPP, the extremum (minimum or maximum) always occurs at one of the corner points of the feasible region. Evaluate z = 3x - 4y at all corner points and identify which gives the smallest value.
<p><strong>Step 1:</strong> Evaluate the objective function z = 3x - 4y at each corner point:</p><p>• At (0, 0): z = 3(0) - 4(0) = <strong>0</strong></p><p>• At (0, 8): z = 3(0) - 4(8) = <strong>-32</strong></p><p>• At (4, 10): z = 3(4) - 4(10) = 12 - 40 = <strong>-28</strong></p><p>• At (6, 8): z = 3(6) - 4(8) = 18 - 32 = <strong>-14</strong></p><p>• At (6, 5): z = 3(6) - 4(5) = 18 - 20 = <strong>-2</strong></p><p>• At (5, 0): z = 3(5) - 4(0) = <strong>15</strong></p><p><strong>Step 2:</strong> Compare all values: 0, -32, -28, -14, -2, 15</p><p>The minimum value is <strong>-32</strong>, which occurs at the point <strong>(0, 8)</strong></p><p>∴ Answer: <strong>D</strong> (minimum occurs at (0, 8))</p>
Correct Answer: D