1. Linear Programming

2. Consider the following problem: A tailor has the following materials available in stock: 16 m 2 of cotton, 11 m 2 of silk and 15 m 2 of wool. A suit requires 2 m 2 of cotton, 1 m 2 of silk and 1 m 2 of wool to produce, while a gown requires 1 m 2 of cotton, 2 m 2 of silk and 3 m 2 of wool. A clothing outlet will pay the tailor $150 for each suit and $250 for each gown he produces. How many suits and how many gowns should the tailor prepare in order to maximize his revenue from the sale of these articles of clothing?

3. First we must consider some mathematical theory and review

4. y = mx + b x y

5. y > mx + b lies above the line x y 1 b = –4 m = 3 test point (0, 0) LS = y = 0 RS = 3(0) – 4 RS = –4 LS > RS  (0, 0) is in the region y ≥ 3x – 4 y < mx + b lies below the line The line is in the region as well RS = 3(x) – 4 –4 3

6. If we solve the inequality for y we get y > mx + b (above the line) x y –2 a = 5 b = –2 test point (0, 0) LS = 0 RS = 10 LS < RS  (0, 0) is in the region 2x – 5y < 10 5 The line is not in the region

7. x > a lies to the right of the line x y 2 a = 2 vertical line test point (0, 0) LS = 0 RS = 6 LS < RS  (0, 0) is not in the region 3x > 6 x < a lies to the left of the line

8. x y –2 x – y  2 2x + y  –2 y  1 2 1 –1 Above this line Below this line Above this line The boundary lines are included

9. y x Point Intercept Position Slope Inclination Direction Consider the straight line on the x-y plane y = mx + b y = m(x  a) Ax + By + C = 0

10. y x The family of lines with slope of 2 y = 2x + b

11. y x The family of lines with y-intercept 2 y = mx + 2 2

12. m is called a parameter (or free variable) y = 3x + b y = a(x – h) 2 + k x 2 + y 2 = r 2 Here are some additional examples: y = asin(x – p) + q y = mx + 2 y = log b x + c

13. y x The parameter k is part of an expression that determines the y-intercept (or the x-intercept). 3x – 4y = k The slope is always ¾

14. Definitions: The objective function: The feasible region: The corner points: A linear function (consisting of two variables if a plane is used) which is to be maximized or minimized. The set of interior and boundary points of a polygon defined by the inequality relations which specify the restrictions (the constraints) on the variables used. The vertices of the feasible region (the set of possible solutions to the problem).

15. The theory of linear programming states that, to find the optimum value of f, where f = Ax + By, we need only test the values of f at the corner points of the feasible region R. This assumes that region R is a convex set of points determined by the intersection of 3 or more linear inequalities. The region R may be open (also called bounded) or closed (also called unbounded). The region is usually open when we are testing for a minimum value of f.

16. C B Q P D O A y x B Maximum The Objective Function

17. C B Q P D O A y x B C Maximums The Objective Function

18. A C B Q P D O y x B Minimum The Objective Function

19. Maximize: 150x + 250yR = Products Resources Suits (x)Gowns (y) Available Cotton2116 Silk1211 Wool1315 Product - Resource Chart (figures in m 2 )

20. Subject to: 2x + 1y  16 1x + 2y  11 1x + 3y  15 x  0 y  0 The quantities must be positive

21. y x 16 15 5 5.5 8 11 A (0, 5) B (3, 4) E (6.6, 2.8) D (8, 0) C (7, 2) O (0, 0)

22. 2x + y = 16 [1] x + 3y = 15 [2] 6x + 3y = 48 [3] 3  [1] [3]  [2] 5x = 33 x = 6.6 y = 2.8Substituting in [1] Divide by 5 Thus, (x, y) = (6.6, 2.8) Point E in previous diagram

23. Conclusion: The maximum revenue can be generated by producing 7 suits and 2 gowns Table for Testing the Revenue Function Corner PointObjective Function: 150x + 250y A ( 0, 5 ) R = 150(0) + 250(5) B ( 3, 4 ) R = 150(3) + 250(4) = 1450 C ( 7, 2 ) R = 150(7) + 250(2) = 1550 D ( 8, 0 ) R = 150(8) + 250(0) = 1200 O (0, 0) R = 150(0) + 250(0) = 0 = 1250 = 1550