An LPP: Minimize z = 30x − 18y subject to 3x + 4y ≤ 60, 5x − 3y ≥ 20, 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)
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Correct answer: A. A and C only
Explanation
z = 30x − 18y = 6(5x − 3y), and the constraint forces 5x − 3y ≥ 20, so the minimum is z = 120, reached at every feasible point on the line 5x − 3y = 20. (4, 0) and (7, 5) lie on it and satisfy 3x + 4y ≤ 60. (2, 0) and (0, 15) are infeasible, and (8, 5) gives z = 150.
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