Unit 1 Linear programming

Define: LINEAR PROGRAMMING – is a method for finding a minimum or maximum value of some quantity, given a set of constraints (Limits). Define: OBJECTIVE FUNCTION – the quantity you are trying to maximize or minimize Define: FEASIBLE REGION – The area created by the system of inequalities (constraints)

Find the Maximum for P = 5x - y Substitute each of the coordinates (A-D)into the equation above to see which gives the maximum value. A: P = 5(0) – 0 P = 0 B: P = 5(6) – 0 P = 30 C: P = 5(2) – 4 P = 6 D: P = 5(0) – 4 P = – 4

Find the Maximum & Minimum for P = -.04x + 3.2y Find the coordinates and then substitute each (A-F)into the equation above to see which gives the maximum and minimum values. A: P = –0.04(0) + 3.2(2) P = 6.4 A (0, 2) B (0, 5) C (1, 6) D (5, 2) E (5, 0) F (4, 0) B: P = –0.04(0) + 3.2(5) P = 16 C: P = –0.04(1) + 3.2(6) P = D: P = –0.04(5) + 3.2(2) P = 6.2 E: P = –0.04(5) + 3.2(0) P = – 0.2 F: P = –0.04(4) + 3.2(0) P = – 0.16 Maximum Minimum

Solving linear programming Minimum for: C = 3x + 4y (2, 1) (4, 1) (2, 2) (2, 1) (2, 2) (4, 1)

You are making H-Dub T-shirts & Hats to sell for homecoming and under the following constraints. You have at most 20 hours to work You want to have at most 50 items to sell H-DUB HAT Takes 10 minutes to make Supplies cost $4 Profit $6 H-DUB T-SHIRT Takes 30 minutes to make Supplies cost $20 Profit $20 Time = 10x + 30y ≤ 1200 Amount= x + y < 50 Real life= x ≥ 0 & y ≥ 0 Profit: P = 6x + 20y Constraints: Objective function: | | | – 100 – 50 – (50, 0) (15, 35) (0, 0) (0, 40) Profit: P = 6x + 20y (0, 0) P = 6(0) + 20(0) P = 0 (50, 0) P = 6(50) + 20(0) P = 300 (15, 35) P = 6(15) + 20(35) P = 790 (0, 40) P = 6(0) + 20(40) P = 800

CLASSWORK P. 160