A linear programming problem (LPP) is as follows: Min z = 30x − 18y, subject to the constraints 3x + 4y ≤ 60, 5x − 3y ≥ 20 and x, y ≥ 0. In this feasible region, the solution(s) of the LPP is/are
A. (4, 0)
B. (2, 0)
C. (7, 5)
D. (0, 15)
E. (8, 5)
Choose the correct answer from the options given below:
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Correct answer: A. A and C only
Explanation
z = 30x − 18y = 6(5x − 3y) ≥ 6 × 20 = 120, so the minimum 120 is reached at every feasible point on the line 5x − 3y = 20. (4, 0) and (7, 5) both lie on it and satisfy 3x + 4y ≤ 60. (2, 0) and (0, 15) are infeasible, and (8, 5) gives z = 150.More Discrete Structures and Optimization questions
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