COMEDK · Maths · 35. Linear Programming
Maximum value of \(z=12 x+3 y\), subject to constraints \(x \geq 0, y \geq 0, x+y \leq 5\) and \(3 x+y \leq 9\) is
- A 15
- B 36
- C 60
- D 40
Answer & Solution
Correct Answer
(B) 36
Step-by-step Solution
Detailed explanation
The objective function is \(z = 12x + 3y\). The constraints are: 1) \(x \geq 0, y \geq 0\) 2) \(x + y \leq 5\) 3) \(3x + y \leq 9\) The feasible region is a polygon with vertices determined by the intersection of the lines: Intersection of \(x+y=5\) and \(3x+y=9\): Subtracting…
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