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Decision Variables: x1 = Variety, x2 = Bakery Supplies Objective Function: Maximize Z = 18000x1+315000x2 Set of Constraints: x1 ≤ 6000, x2 ≤ 315000, x1 + x2 ≤ 48000, x1 + x2 > 3,230 Parameters: Php 6,000 for Variety, Php 315,000 for Bakery Supplies, Php 18,000 for Variety, Php 48,000 for Bakery Supplies, Php 3,230 for Other Equipment. For Graphical Solution Graph the Constraints: x1 + x2 < 48000 x2 < 315000 x1 ≤ 6000 x1 + x2 ≥ 3,230 Obtain the Corner-Points of the Feasible Solution: (0, 0), (6000, 315000), (48000, 0), (3,230, 44870) Get the Optimal Solution: The optimal solution is (6000, 315000) with a maximum profit of Php 4,155,000.

Decision Variables: x1 = Variety, x2 = Bakery Supplies Objective Function: Maximize Z = 18000x1+315000x2 Set of Constraints: x1 ≤ 6000, x2 ≤ 315000, x1 + x2 ≤ 48000, x1 + x2 > 3,230 Parameters: Php 6,000 for Variety, Php 315,000 for Bakery Supplies, Php 18,000 for Variety, Php 48,000 for Bakery Supplies, Php 3,230 for Other Equipment. For Graphical Solution Graph the Constraints: x1 + x2 < 48000 x2 < 315000 x1 ≤ 6000 x1 + x2 ≥ 3,230 Obtain the Corner-Points of the Feasible Solution: (0, 0), (6000, 315000), (48000, 0), (3,230, 44870) Get the Optimal Solution: The optimal solution is (6000, 315000) with a maximum profit of Php 4,155,000.

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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Decision Variables: x1= Variety, x2 = Bakery Supplies Objective Function: Maximize Z=18000x1+315000x2 Set of Constraints: x1 ≤ 6000, x2 ≤ 315000, x1 + x2 < 48000, x1 + x2 ≥ 3,230 Parameters: Php 6,000 for Variety, Php 315,000 for Bakery Supplies, Php 18,000 for Variety, Php 48,000 for Bakery Supplies, Php 3,230 for Other Equipment. For Graphical Solution Graph the Constraints: x1 + x2 < 48000 x2 ≤ 315000 x1 ≤ 6000 x1 + x2 ≥ 3,230 Obtain the Corner-Points of the Feasible Solution: (0, 0), (6000, 315000), (48000, 0), (3,230, 44870) Get the Optimal Solution: The optimal solution is (6000, 315000) with a maximum profit of Php 4,155,000.
Transcribed Image Text:Decision Variables: x1= Variety, x2 = Bakery Supplies Objective Function: Maximize Z=18000x1+315000x2 Set of Constraints: x1 ≤ 6000, x2 ≤ 315000, x1 + x2 < 48000, x1 + x2 ≥ 3,230 Parameters: Php 6,000 for Variety, Php 315,000 for Bakery Supplies, Php 18,000 for Variety, Php 48,000 for Bakery Supplies, Php 3,230 for Other Equipment. For Graphical Solution Graph the Constraints: x1 + x2 < 48000 x2 ≤ 315000 x1 ≤ 6000 x1 + x2 ≥ 3,230 Obtain the Corner-Points of the Feasible Solution: (0, 0), (6000, 315000), (48000, 0), (3,230, 44870) Get the Optimal Solution: The optimal solution is (6000, 315000) with a maximum profit of Php 4,155,000.
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