MATH 1324 (Finite Mathematics or Business Math I) Lecture Notes Author / Copyright: Kevin Pinegar

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1 MATH 34 Module 3 Notes: SYSTEMS OF INEQUALITIES & LINEAR PROGRAMMING 3.3 SOLVING LINEAR PROGRAMMING PROBLEMS GEOMETRICALLY WE WILL BEGIN THIS SECTION WITH AN APPLICATION The Southern States Ring Company designs and sells two types of rings: the brass and the aluminum. They can produce up to 4 rings each day using up to 60 total man-hours of labor per day. It takes 3 man-hours to make one brass ring and man-hours to make one aluminum ring. How many of each type of ring should be made daily to maximize the company's profit, if the profit on a brass ring is $40 and on an aluminum ring is $35? Some like to set up a table first. Let x = number of brass rings and y= number of aluminum rings Brass Aluminum Maximum Rings x y 4 Rings Hours 3x y 60 Hours x+y 4 (total number of rings is less than or equal 4) 3x+y 60 (total man-hours is less than or equal 60) Also: x,y 0 (cannot produce a negative number of rings) The objective function is the function we wish to maximize (or minimize). In this case it is profit (P). P=40x+35y Put it all together to get a linear programming model. Maximize : P 40 x 35 y Subject To : x y 4 3 x y 60 x, y 0 After we get the model, we must graph the feasible region. Graph x+y 4 Inequality and 3x+y 60 Boundary x+y=4 a) x y 4 3x+y=60 b) 3x y 60 a) x y 4 and 3x y 60 The dark green area is the overlap (0,4), (4,0) Test (0,0) & Conclusion 0 4, True (0,30), (0,0) 0 60, True Now we will restrict the region to the first quadrant and only list the corner points of the region. The corner points are (0,0),(0,4),(,),(0,0)

2 Why Check Only The Corner Points? Let me show you the answer by looking at the objective function. 40 x 35y P (It s reversed for convenience. Write it slope-intercept form.) 35 y 40x P y 40 x P y 8 x P 7 35 The linear equation above has a slope of m=-8/7 and y-intercept of P/35. Let s try some different profits. For convenience I ll use values divisible by 35. If P=400, x x y. Look at the graph again. The dashed line in the graph is the objective function for P=400. Notice it has no points in the feasible region. Let s try a lower profit. If P=700, x x 0 7 y. Look at the graph again with this line. The dashed line in the graph is the objective function for P=700. Notice it has points in the feasible region, but we could lift the line higher to get a larger profit. Since I know the answer, let s try that. If P= x 7 35 y. The dashed line now is as high as it can go for profit and still touch the region. Did you notice it first touches a corner point (,)? If the values $40 and $35 were different numbers, it might first touch the graph at a different point, but it would still have to be at a corner point like (0,0) or (0,4). Yes! It is possible that you are better off not producing one type of ring. (Depends on the coefficients of the profit function.) Now I will check each corner point to prove it. (0,0), P=40(0)+35(0)=$0 Not much profit. (0,4), P=40(0)+35(4)= $840 (,), P=40()+35()=$900 (0,0), P=40(0)+35(0)=$800 Optimal MAX P=$900 at x=, y=

3 Guidelines For Solving Linear Programming Problems. Setup the linear programming model if a word problem. a. Make sure to first list the objective function b. Followed by the problem constraints c. Last; make sure you write the nonnegative constraints. Graph the feasible region from the set of inequalities (as long as you have nonnegative constraints, you can focus your graph on the first quadrant.) a. If the feasible region is bounded, there exists a maximum and a minimum. b. If the feasible region is unbounded, only a minimum exists. c. If there is no feasible region (no overlap), then there is neither a maximum, nor a minimum. 3. Assuming there is a solution; Identify all corner points of the feasible region. a. Some corner points are found on the x or y axis. b. Other corner points are found where the constraints intersect. c. The origin (0,0), may or may not be a corner point. 4. Substitute each corner point into the objective function. a. If the objective function is to be maximized, the maximum is the largest value of the objective function. b. If the objective function is to be minimized, the maximum is the smallest value of the objective function. Example : Maximize: P 50x 80y Subject To : x y 3 3x 4 y 84 x, y 0 Graph of feasible region: The feasible region is the darker area below, bounded by the four corner points listed. The point (0,6) was found by solving the system, x+y=3 3x+4y=84 Boundary Test (0,0) & Conclusion x y 3 (0,6), (3,0) 0 3, True 3x 4y 84 (0,), (8,0) 0 84, True The region is bounded, so a maximum exists. P=50x+80y (0,0) P=0 (0,6) P=80 (0,6) P=480 (8,0) P=400 The maximum is: P=480 at x=0, y=6 Warning: The optimal solution is not always inside the first quadrant. Suppose P=0x+80y P=0x+80y (0,0) P=0 (0,6) P=80 (0,6) P=880 (8,0) P=560 Solution is now P=80 at x=0, y=6 3

4 Example : C 5 x x x 3 x 5 Subject To : x x 0 x, x 0 Boundary x 3x 5 x x 0 Graph of feasible region: The feasible region is the darker area below, bounded by the three corner points listed. The point (9,) was found by solving the system, x 3x 5 x x 0 (0,5), (5,0) Test (0,0) & Conclusion 0 5, False (0,), (8,0) 0 0, False The region is unbounded, so a minimum exists, but no maximum. C=5x+x (0,0) C=40 (9,) C=49 (5,0) C=75 The minimum is: C=40 at x=0, x=0 Warning: The optimal solution is not always inside the first quadrant. Suppose C=5x+0x C=5x+0x CP=400 (0,0) CP=85 (9,) C (5,0) P=75 C Solution is now P=75 at x=5, y=0 NOTE: Neither function would have a maximum, since the region is unbounded. Practice Problems (Answers Given) Maximize : Z 5 x 3 x x x 0 Subject To : x x 8 x, x 0 Z 3 x 7 x x x 0 Subject To : x x 8 x, x 0 (0,0) (0,4) (4,) (5,0) Z=5x+3x ZP=0 ZP= ZP=6 ZP=5 Maximum: Z=6, at (4,) (0,0) (4,) (8,0) Z=3x+7x ZP=70 P=6 Z ZP=4 Minimum: Z=4, at (8,0) 4

5 Example 3 Maximize/ Z 3x 4x x 4x 4 3x 3x Subject To : 4x x 0 x, x 0 The feasible region is the darker triangular region bounded by points (5,0),(3,4),(7,0). Boundaries Testing Conclusion x +4x =4 (0,6), (,0) 04 True 3x +3x = (0,7), (7,0) 0 True 4x +x =0 (0,0), (5,0) 00 False Solutions: Z=3x+4x (5,0) Z=5 (3,4) Z=5 (7,0) Z= Since region is bounded, it has both a maximum and minimum, Maximum: Z=5 at (3,4) Minimum: Z=5 at (5,0) Practice Problem (Answer Given) Maximize/ Z 3x x x x 0 0x x 36 Subject To : x 5x 36 x, x 0 Maximize/ Z 0x 6x x 36 x 4x 3 Subject To : x 0 x, x 0 0x Z=3x+x (3,6) Z=5 (,6) Z= (8,4) Z=8 Maximum: Z=8, at (8,4) Minimum: Z=5 at (3,6) Z=0x+0x (0,0) Z=400 (0,8) Z=360 (4,6) Z=60 (6,0) Z=60 Minimum: Z=60 at (4,6) or (6,0) The function is minimized at either of the two points. There is no maximum since the region is unbounded. Tried to trick us, yes? 5

6 Applications of Linear Programming Problems ) Medicine: A patient in a hospital is required to have at least 84 units of drug A and 0 units of drug B each day (we will assume an overdose is harmless). Each gram of substance M contains 0 units of drug A and 8 units of drug B. Each gram of substance N contains units of drug A and 4 units of drug B. Now suppose that both M and N contain a slightly harmful drug W. There is 3 units of drug W in each gram of drug M and unit of drug W in drug N. How many grams of substance M and N can be mixed to meet the minimum daily requirements of drugs M and N and at the same time, minimize the amount of drug W. We set the constraints up in an earlier section. Let x= number of grams of substance M and y be the number of grams of substance N. System of Inequalities 0x y 84 8x 4 y 0 x, y 0 Now we must introduce objective function W. The total dosage for drug W can be represented by W=3x+y. So our linear programming model becomes, W 3x y 0x y 84 Boundaries Testing Conclusion 0x+4y=84 (0,4), (8.4,0) 0 84 False 8x+4y=0 (0,30), (5,0) 0 0 False Subject To8x 4 y 0 x, y 0 The feasible region is the darker unbounded region region bounded by points with corner points (0,4),(4,),(5,0). Solutions: Z=3x+y (0,4) Z=4 (4,) Z=34 (5,0) Z=45 Since region is unbounded, it only has both a minimum, Minimum: Z=34 at (4,) So you should administer 4 units of drug M and units of drug N to minimize drug W and satisfy the constraints. 6

7 . The following is an example of a standard minimization problem. In later sections we will solve these problems using the DUAL method. A manufacturer of dogfood makes two secret ingredients that goes into their dogfood, codenamed: Superdog and Underdog. Each kg of Superdog contains 300 grams of vitamins, 400 grams of protein, and 00 grams of carbohydrate. Each kg of Underdog contains 00 grams of vitamins, 300 grams of protein, and 00 grams of carbohydrate. Minimum nutritional guidelines require that a mixture made from these ingredients contain at least 900 grams of vitamins, 00 grams of protein, and 800 grams of carbohydrate. Superdog costs $.00 per kg to produce and Underdog costs $.5 per kg to produce. Find the number of kilograms of each ingredient that should be produced in order to minimize cost. {let x=the # of kg of Superdog, and x=the # of kg of Underdog} i) Set up the problem ii)solve the problem by graphing. Minimize : C x.5 x vit. 300 x 00 x x 300 x 00 pro. ST : carb. 00 x 00 x 800 x, x 0 nonneg. (0,9) (,6) (4,) (8,0) Graphing The Feasible Region Boundaries Testing Conclusion 300x+00x=900 (0,9), (3,0) False Points C=x+.5x 400x+300x=00 (0,/3), (/,0) False 0 00 (0,9) C=.5 00x+00x=800 (0,4), (8,0) False (,6) C=9.50 The feasible region is the graph on the right with corner points (4,) C=8.50 (0,9),(,6), (4,), (8,0). (8,0) C=6.00 Minimum is C=$8.50 for x= 4 kg of Super, x= kg of Under 3. The following is an example of a standard MAXimization problem. In later sections we will solve these problems using the SIMPLEX method. A chemical firm makes two types of industrial solvents, S and S. Each solvent is a mixture of three chemicals. Each kl of S requires L of chemical A, 9L of chemical B, and 30L of chemical C. Each kl of S requires 4L of chemical A, 5L of chemical B, and 30L of chemical C. The profit per kl of S is $00, and the profit per kl of S is $85. The inventory of the company shows 480 L of chemical A, 80 L of chemical B, and 70 L of chemical C. Assuming the company can sell all the solvent it makes, find the number of kl of each solvent that the company should make to maximize profit. {let x=the # of kl of S, and x=the # of kl of S} Maximize : P 00 x 85 x A x 4 x 480 9x 5x 80 B ST : C 30 x 30 x 70 x, x 0 nonneg. Graphing The Feasible Region Boundaries Testing Conclusion x+4x=480 (0,0), (40,0) True x+5x=80 (0,36), (0,0) True x+30x=70 (0,4), (4,0) True 0 70 The feasible region is the graph on the right with corner points (0,0), (0,0), (8,6), (5,9), (0,0). Maximum is C=$65 for x= 5, x=9 kg of Underdog (0,0) (8,6) (5,9) (0,0) Points (0,0) (0,0) (8,6) (5,9) (0,0) P=00x+85x P=0 P=700 P=60 P=65 P=000 7