Linear Programming Problems: Geometric Solutions

Transcription

1 Linear Programming Problems: Geometric s Terminology Linear programming problems: problems where we must find the optimum (minimum or maximum) value of a function, subject to certain restrictions. Objective function: the function we are trying to optimize. Constraints: the set of restrictions, given by linear inequalities, which the solution must satisfy. Non-negativity constraints: x 0, y 0 These constraints limit the feasible region to the first quadrant. Feasible Region: the area where all of the inequalities are simultaneously true. A feasible region which extends indefinitely in one direction is called an unbounded region. A bounded region is hemmed in on all sides by borders. Corner points: the points where the borders of the feasible region cross. To find the corner points of the feasible region you may need to solve a system of linear equations. To do this, you may use the substitution method, the elimination method, or the rref program. Corner Point Theorem If an optimum value (min or max) of the objective function exists, it will occur at one or more of the corner points of the feasible region or on the line segment between two corner points. Example 1: The following graph shows a region of feasible solutions. Use this region to find maximum and minimum values of the given objective functions.

2 a. z =.40x +.75y b. z = 1.50x +.25y Evaluate each objective function at each corner point. The table below shows the function values. The minimum value of z1 = 0. It occurs when x = 0 and y = 0. (Note: This is often a trivial solution.) The maximum value of z1 = 9. It occurs when x = 0 and y = 12. The minimum value of z2 = 0. It occurs when x- 0 and y = 0. (Note: This is often a trivial solution.) The maximum value of z2 = 12. It occurs when x = 8 and y = 0. How to Solve a Linear Programming Problem Graphically 1. If necessary, identify the unknowns, label the variables, write the objective function, and write the constraints. 2. Graph the system of constraints. Find the feasible region and the corner points. 3. Find the value of the objective function for each corner point. Identify the minimum and / or maximum value of the objective function and the point (or points) where this optimum value occurs.

3 Example 2: Use the graphical method to solve the following linear programming problem. Maximize z = 3x + y subject to x 1, y 0, 2x + y 6. Graph the feasible region: The feasible region is a bounded region with 3 corner points. Evaluate the objective function at each corner point. corner points objective function values (1, 0) z(1, 0) = 3(1) + 0 =3 (min) (1, 4) z(1, 4) = 3(1) + 4 = 7 (3, 0) z(3, 0) = 3(3) + 0 = 9 (max) : A maximum value of z = 9 occurs at (3, 0) and a minimum value of z = 3 occurs at (1, 0). How to Solve a Linear Programming Application Problem 1. Read the problem carefully and identify the unknowns. Assign each unknown a variable. 2. If possible, organize the given information into a table. Each column of the table will be labeled with one variable and each row will be labeled with a constrained element. 3. From your table, write the objective function and all necessary constraints. [Hint: Don t forget the non-negativity constraints, i.e. x 0, y 0.] 4. Graph the system of constraints. Find the feasible region and the corner points. 5. Find the value of the objective function for each corner point. Identify the minimum and / or maximum value of the objective function and the point where this optimum value occurs.

4 Example 3: Suppose a sporting goods factory can turn out 5000 balls each week. To meet the needs of its regular customers, the factory must produce 1500 basketballs and 2000 baseballs. If the profit for each basketball is $15 and the profit for each baseball is $5, how many of each type of ball should the factory produce to maximize profit? Define variables. Let x = the number of basketballs produced. Let y = the number of baseballs produced. Write inequalities. x 1500 y 2000 x + y 5000 Graph the system. The vertices are (1500, 2000), (3000, 2000), and (1500, 3500). Write an expression. Since profit is $15 per ball for basketballs and $5 per ball for baseballs, the profit function is P(x, y) = 15x + 5y. Substitute values. P(1500, 2000) = 15(1500) + 5(2000) = 32,500 P(3000, 2000) = 15(3000) + 5(2000)= 55,000 P(1500, 3500) = 15(1500) + 5(3500) = 40,000 Answer the problem. The maximum profit occurs when 3000 basketballs and 2000 baseballs are produced.

5 Example 4: The Happy Times Toy Shop makes wooden cars and wooden tops. Each car requires 5 hours of woodworking and 2 hours of painting. Each top requires 3 hours of woodworking and 1 hour of painting. The profit is $10 on each car and $6 on each top. There are 60 hours available each week for woodworking and 35 hours available for painting. How many of each item should be produced in order to maximize profit? Define variables. Let c = the number of cars produced. Let t = the number of tops produced. Write inequalities. c 0 t 0 5c + 3t 60 No more than 60 hours woodworking. 2c + t 35 No more than 35 hours painting. Graph the system. The vertices are (0, 0), (12, 0), and (0, 20). Write an expression. Since profit on each car is $10 and the profit on each top is $6, the profit function is P(c, t) = 10c + 6t. Substitute values. P(0, 0) = 10(0) + 6(0) or 0 P(12, 0) = 10(12) + 6(0) or 120 P(0, 20) = 10(0) + 6(20) or 120 Answer the problem. The problem has alternate optimal solutions. The shop will make the same profit if they produce 12 cars and 0 tops as it will from producing 0 cars and 6 tops.