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Linear Programming Model: Subject to: 2. Linear Programming Model: Minimize Cost = 50,000X + 75,000Y 10X + 30Y ≥ 60 X ≤ 10,000 5X + 10Y ≥ 25 The optimal solution is X = 2 hectares of kamagong and Y =2 hectares of narra, with a minimum cost of P200,000. 6X +3Y ≥ 12 X,Y ≥ 0 Maximize Income = 0.035(50,000 - X) + 0.045X Subject to: X ≥ 30,000 The optimal solution is X = 30,000, with a maximum monthly income of P1,650. ARE NUMBERS 1 & 2 CORRECT? SHOW THE GRAPH WITH LABELS AND SOLVE BOTH

Linear Programming Model: Subject to: 2. Linear Programming Model: Minimize Cost = 50,000X + 75,000Y 10X + 30Y ≥ 60 X ≤ 10,000 5X + 10Y ≥ 25 The optimal solution is X = 2 hectares of kamagong and Y =2 hectares of narra, with a minimum cost of P200,000. 6X +3Y ≥ 12 X,Y ≥ 0 Maximize Income = 0.035(50,000 - X) + 0.045X Subject to: X ≥ 30,000 The optimal solution is X = 30,000, with a maximum monthly income of P1,650. ARE NUMBERS 1 & 2 CORRECT? SHOW THE GRAPH WITH LABELS AND SOLVE BOTH

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Linear Programming Model: Subject to: 2. Linear Programming Model: Subject to: Minimize Cost = 50,000X + 75,000Y The optimal solution is X = 2 hectares of kamagong and Y =2 hectares of narra, with a minimum cost of P200,000. X ≥ 30,000 10X + 30Y ≥ 60 X ≤ 10,000 5X + 10Y ≥ 25 6X + 3Y ≥ 12 Maximize Income = 0.035(50,000 - X) + 0.045X X,Y ≥ 0 The optimal solution is X = 30,000, with a maximum monthly income of P1,650. ARE NUMBERS 1 & 2 CORRECT? SHOW THE GRAPH WITH LABELS AND SOLVE BOTH
Transcribed Image Text:Linear Programming Model: Subject to: 2. Linear Programming Model: Subject to: Minimize Cost = 50,000X + 75,000Y The optimal solution is X = 2 hectares of kamagong and Y =2 hectares of narra, with a minimum cost of P200,000. X ≥ 30,000 10X + 30Y ≥ 60 X ≤ 10,000 5X + 10Y ≥ 25 6X + 3Y ≥ 12 Maximize Income = 0.035(50,000 - X) + 0.045X X,Y ≥ 0 The optimal solution is X = 30,000, with a maximum monthly income of P1,650. ARE NUMBERS 1 & 2 CORRECT? SHOW THE GRAPH WITH LABELS AND SOLVE BOTH
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