x Boundary Intercepts Test (0,0) Conclusion 2x+3y=12 (0,4), (6,0) 0>12 False 2x-y=2 (0,-2), (1,0) 0<2 True

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1 MATH 34 (Finite Mathematics or Business Math I) Lecture Notes MATH 34 Module 3 Notes: SYSTEMS OF INEQUALITIES & LINEAR PROGRAMMING 3. GRAPHING SYSTEMS OF INEQUALITIES Simple Systems of Linear Inequalities We will start out small and work our way up the chain. Below are a couple of systems of linear inequalities. Guidelines For Solving A System Of Linear Inequalities. Graph each inequality using the methods from the previous section.. Shade each region with a different color pencil. (map pencils are good) 3. Identify the overlapping region and shade that region darker than the others. That region is called the feasible region. (I will also identify the vertices of the region, because this is important for solving linear programming problems. The vertices are called corner points.) 3y ) x y x Boundary Intercepts Test (0,0) Conclusion x+3y= (0,4), (6,0) 0> False x-y= (0,-), (,0) 0< True x+3y > x- y System Unbounded region since it goes on and on. Summary: The feasible region is the darker green area. All points within that region are solutions to both inequalities. This is a visual representation of the solution set. Where the two lines intersect is called a corner point. Two find the corner point, solve the system: x 3y x y You can solve this system using matrices or the more primitive addition method. The solution is (9/4,5/) = (.5,.5)

2 y 0 ) x y 8 MATH 34 (Finite Mathematics or Business Math I) Lecture Notes x Boundary Intercepts Test (0,0) Conclusion x+y=0 (0,0), (5,0) 0 0 True x+y 0 x+y=8 (0,4), (8,0) 0 8 True x+y 8 System Unbounded region since it goes on and on. Summary: The feasible region is the darker green area. All points within that region are solutions to both inequalities. This is a visual representation of the solution set. Where the two lines intersect is called a corner point. Two find the corner point, solve the system: x y 0 x y 8 You can solve this system using matrices or the more primitive addition method. The solution is (4,). This the corner point. TEACHER S NOTE: When you graph a system of linear inequalities, make sure you emphasize the overlapping area. Definition of Corner Point: A corner point of a solution region is a point in the solution region that is the intersection of two boundary lines. (It must be within the boundary or on the boundary.) Notice the shaded region to the right. The arrows are pointing to the corner points. There is one other corner point. Do you see it? It is the origin (0,0). The list of corner points are: (0,0), (0,0), (5,8), (9,4), (,0)

3 More Complicated System 3) 3x 4x x 0 MATH 34 (Finite Mathematics or Business Math I) Lecture Notes x 4x 4 System Boundaries Intercepts Testing Conclusion x +4x 4 x +4x =4 (0,6), (,0) 04 True 3x 3x +3x 3x +3x = (0,7), (7,0) 0 True 4x +x 0 4x +x =0 (0,0), (5,0) 00 False ) x+4x4 ) x+4x4 and 3x+3x 3) 4x+x 0 Now let s inlay #3 on # to complete our graph. The feasible region is the darkest purple region that continues to go down, down, down. There is only one corner point (3,4), where the lines 3x+3x and 4x+x 0 intersect. 3

4 MATH 34 (Finite Mathematics or Business Math I) Lecture Notes Systems of Linear Inequalities Restricted to the First Quadrant 4) y x y 3 x 5y 50 x, y 0 x System Boundaries Intercepts Testing (0,0) Conclusion x+y x+y = (0,), (,0) 04 True x+y 3 x+y =3 (0,3), (3,0) 03 True x+5y50 x+5y=50 (0,0), (5,0) 050 True ) x+y ) x+y and x+y 3 3) x+y, x+y 3, x+5y50 4) x+y, x+y 3, x+5y50, x 0, y 0 The last two constraints restrict our feasible region to the first quadrant giving us the picture below. Note: At this point the region is the darkest purple region that is unbounded. (5,0) is not in the region. I left it visible because it was one of the intercepts of the last boundary I added. We have not limited the region to the first quadrant yet. That comes in the next picture. Notice the region has 5 corner points. (0,0), (0,0),(5,8),(9,4),(,0) The points (5,8),(9,4) are results of solving the systems of lines that give those points. 4

5 Practice Problem: x 3x 4x 4x 3x x x, y (Solution at the right) MATH 34 (Finite Mathematics or Business Math I) Lecture Notes Solution should be the triangle bounded within corner points (3,4), (5,0), (7,0) A Note About Linear Programming: Linear Programming is a process where a function is optimized based on the points restricted to a specific region. For example, suppose we needed to Maximize a revenue function R=50x+00x ; where x represent 3 speed bikes and x represents 5 speed bikes and we could only use values within the above triangular region. The region contains many points. It turns out that the maximum must occur at one of the corner points. So, we can plug each point in and see which gives the maximum. Corner Point R=50x+00x (5,0) R=$750 (7,0) R=$050 (3,4) R=$50 (optimal solution) We could also minimize a function using this method. Note that the minimum is R=$750 We would need to study further to understand this topic. Application: 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. How many grams of substance M and N can be mixed to meet the minimum daily requirements of each drug? Solution: Let x= number of grams of substance M and y be the number of grams of substance N. Total of Inequality Minimum System of Inequalities Each Drug ( ) Requirements 0x y 84 0x+y 84 Drug A 8x+4y 0 Drug B 8x 4 y 0 x and y must also be greater than or equal to zero. We cannot mix a negative amount of M or N. x, y 0 5

6 MATH 34 (Finite Mathematics or Business Math I) Lecture Notes Linear Programming: Linear programming is optimizing an objective function while given certain limiting conditions {constraints}. We will learn several methods to solve a linear programming {LP} problem in this chapter. The first such method is the graphing or geometric method. Solving Linear Programming Problems Geometrically: This method is best when there are only two variables. It is very difficult to graph inequalities in 3 dimensions and you cannot graph in 4 dimensions. We will learn another method later that will allow us to solve LP problems of more than variables. Fundamental Theorem of Linear Programming: If a linear programming problem has an optimal solution, it must occur at one or more corner points. {This means that we only have to check the corner points for the optimal solution.} Properties of Linear Programming Problems: ) If the region is bounded, there exists a maximum and a minimum. a. Consider the following request: Loan me at least $ but no more than $. This is bounded. The minimum loan is $ and the maximum loan is $. ) If the region is unbounded, there exists a minimum but no maximum. a. Consider the following request: Loan me at least $. This is unbounded. The minimum loan is $ but no maximum exists. 3) If the region has no solution, there is neither a maximum nor a minimum. a. Consider the following request: Loan me at least $ but no more than $. This has no solution, therefore there can be no min nor max. Identifying the parts of a LP problem. Maximize: P Subject To : 5x x x x x x, x 3x x and x are called decision variables. Ob jective Function constraint constraint nonnegative constraints Note: In this chapter, you need to adapt to the use of x and x. Just treat x as x and x as y. Steps in solving an LP problem, geometrically see next section ) Graph the region (feasible region). ) Identify the corner points. 3) Evaluate the objective function at each corner point. 4) Interpret the optimal solution {maximum or minimum and where it occurs} We will solve linear programming problems in the next section. 6