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Solve graphically the given linear programming problem (minimization problem) Minimize Z = 3a + 5b S.T -3a + 4b ≤ 12 2a-1b = -2 2a + 3b ≥ 12 1a + 0b ≥ 4 0a + 1b ≥ 2 and both a and b are ≥ 0

Solve graphically the given linear programming problem (minimization problem) Minimize Z = 3a + 5b S.T -3a + 4b ≤ 12 2a-1b = -2 2a + 3b ≥ 12 1a + 0b ≥ 4 0a + 1b ≥ 2 and both a and b are ≥ 0

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Solve graphically the given linear programming problem (minimization problem) Minimize Z = 3a + 5b S. T -3a + 4b ≤ 12 2a - 1b = -2 2a + 3b ≥ 12 1a + 0b ≥ 4 0a + 1b ≥ 2 and both a and bare ≥ 0
Transcribed Image Text:Solve graphically the given linear programming problem (minimization problem) Minimize Z = 3a + 5b S. T -3a + 4b ≤ 12 2a - 1b = -2 2a + 3b ≥ 12 1a + 0b ≥ 4 0a + 1b ≥ 2 and both a and bare ≥ 0
Points to be Noted: (1) In inequality -3a + 4b ≤ 12, product/the candidate/activity requires -3 units of the resource. It does not give any meaning (or by manufacturing the product A the manufacturer can save 3 units of resource No.1 or one has to consume -3 units of A. (All these do not give any meaning as far as the practical problems or real world problems are concerned). (ii) In the second inequality, on the right hand side we have -2. This means that -2 units of resource is available. It is absolutely wrong. Hence in solving a l.p.p. problem, one must see that the right hand side we must have always a positive integer. Hence the inequality is to be multiplied by -1 so that the inequality sign also changes. In the present case it becomes: -2a + 1b ≤ 2.
Transcribed Image Text:Points to be Noted: (1) In inequality -3a + 4b ≤ 12, product/the candidate/activity requires -3 units of the resource. It does not give any meaning (or by manufacturing the product A the manufacturer can save 3 units of resource No.1 or one has to consume -3 units of A. (All these do not give any meaning as far as the practical problems or real world problems are concerned). (ii) In the second inequality, on the right hand side we have -2. This means that -2 units of resource is available. It is absolutely wrong. Hence in solving a l.p.p. problem, one must see that the right hand side we must have always a positive integer. Hence the inequality is to be multiplied by -1 so that the inequality sign also changes. In the present case it becomes: -2a + 1b ≤ 2.
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