Chapter 3 Section 1 A Linear Programming Problem

Read the Following (Very Slowly and Carefully) 1. The chapter 3 introduction paragraph (before section 3. 1 starts) on page 117 2. The steps to linear programming in the blue-grey boxes on pages 125 and 126 3. Section 3. 1 (pages 117– 120)

Linear Programming - Step 1 (Modified) (Modified from the Blue-grey Box on Page 125) • Translate the problem into mathematical language: – Read problem and determine the question being asked. – Define the variables – Form the objective function – Create a table – Form the linear inequalities that restrict the values of the variables.

Linear Programming - Step 2 (Blue-grey Box on Page 126) • Graph the feasible set – Put the inequalities into standard form – Graph the line for each of the inequalities – For each inequality, cross off the portion of the plane that does not satisfy the inequality

Linear Programming - Step 3 (Blue-grey Box on Page 126) • Determine the vertices of the feasible set

Linear Programming - Step 4 (Modified) (Slightly modified from the Blue-grey box on page 126) • Evaluate the objective function at each of the vertices, determine the optimal point, and answer the question

Objective Function • Definition: An Objective Function is a function (linear in our case) formed by several variable for an aspect of the problem. • Two Examples: 1. Profit = 30 x + 40 y 2. Cost = 20 x + 18. 60 y

Objective of Linear Programming To maximize, or minimize, the objective function

Restrictions • The problem that occurs is that there are restrictions to what values the variables in the objective function can be. • Linear programming is the method to find all the possible values that the variables can be (while meeting the all of the requirements of the restrictions placed on the objective function), and then which of these values maximize or minimize the objective function

Optimize the Objective Function To find the values of the variables (usually x and y) that maximizes (or minimizes) the objective function while satisfying all the restrictions imposed by a system of linear inequalities.

Exercise 5 (Page 121) A truck traveling from New York to Baltimore is to be loaded with two types of cargo. Each crate of cargo A is 4 cubic feet in volume, weighs 100 pounds, and earns $13 for the driver. Each crate of cargo B is 3 cubic feet in volume, weighs 200 pounds, and earns the driver $9. The truck can carry no more than 300 cubic feet of crates and no more than 10, 000 pounds. Also, the number of crates of cargo B must be less than or equal to twice the number of cargo A. How many crates of each type of cargo can be loaded into the truck as to maximize the driver’s earnings?

Define the Variables Being Used Start by defining the two variable (which are actually defined in part (b). • Let x represent the number of crates of cargo A • Let y represent the number of crates of cargo B Note how the variables are defined precisely!

Exercise 5 (page 128) Part (a) Cargo A Volume Weight Earnings Cargo B Total Capacity

Exercise 5 Part (a) Each crate of cargo A is 4 ft 3 in volume, weighs 100 lbs, and earns the $13 for the driver. Cargo A Volume 4 ft 3/crate Weight 100 lbs/crate Earnings $13 / crate Cargo B Total Capacity

Exercise 5 Part (a) (cont. ) Each crate of cargo B is 3 ft 3 in volume, weighs 200 lbs, and earns the $9 for the driver. Cargo A Cargo B Volume 4 ft 3/crate 3 ft 3/crate Weight 100 lbs/crate 200 lbs/crate $13 / crate $9 / crate Earnings Total Capacity

Exercise 5 Part (a) (cont. ) The truck can carry no more than 300 ft 3 in volume and no more than 10, 000 lbs. Cargo A Cargo B Volume 4 ft 3/crate 3 ft 3/crate Weight 100 lbs/crate 200 lbs/crate $13 / crate $9 / crate Earnings Total Capacity 300 ft 3 10, 000 lbs

Exercise 5 Part (a) Answer Cargo A Cargo B Volume 4 ft 3/crate 3 ft 3/crate Weight 100 lbs/crate 200 lbs/crate $13 / crate $9 / crate Earnings Total Capacity 300 ft 3 10, 000 lbs

Exercise 5 Part (b) • Let x be the number of crates of cargo A and y be the number of crates of cargo B. Give the two inequalities that x and y must satisfy because of the trucks capacity • Solution: Volume Inequality: 4 x + 3 y < 300 Weight Inequality: 100 x + 200 y < 10, 000

Exercise 5 Part (c) • Give the inequality that x and y must satisfy because of (1) the last sentence of the problem: “Also, the number of crates of cargo B must be less than or equal to twice the number of cargo A. ” and (2) because x and y cannot be negative • Solution: (1) (2) y<2 x x>0 y>0

Exercise 5 Part (d) • Express the earnings from carrying x crates of cargo A and y crates of cargo B ( i. e. State the Objective Function ) • Solution: Earnings = 13 x + 9 y (Note: In the modified version of the steps, this is would occur after the definitions of the variables)

Exercise 5 Part (e) • Graph the feasible set for the shipping problem • Solution: 4 x + 3 y < 300 y < – 4/3 x + 100 x + 200 y < 10, 000 y<2 x y>0 x>0 y < – ½ x + 50 y<2 x y>0 x>0

Exercise 9 (not stated in textbook) • How many pounds of Alfalfa and Corn must the farmer purchase to meet the nutrient needs of his/her cows at the least cost? • Start by defining the two variable Let x represent the number of pounds of Alfalfa Let y represent the number of pounds of Corn

Exercise 9 (pages 122) Part (a) Alfalfa Protein TDN Vitamin A Cost / lbs Corn Requirements

Exercise 9 Part (a) Answer: Protein TDN Vitamin A Cost / lbs Alfalfa Corn Requirements 0. 13 0. 065 4, 500 Prot. lbs/ Alfa. lbs Prot. lbs/ corn lbs Prot. lbs 0. 48 0. 96 26, 880 T. D. N. lbs/ Alfa. lbs T. D. N. lbs/ corn lbs T. D. N. lbs 2. 16 0. 0 43, 200 I. U. ’s/ Alfa. lbs I. U. ’s/ corn lbs I. U. ’s $0. 016

Exercise 9 Part (b) • Let x be the number of pounds of alfalfa hay and y be the number of pounds of corn. Give the inequalities that x and y must satisfy • Solution: 0. 13 x + 0. 065 y > 4, 550 0. 48 x + 0. 96 y > 26, 880 2. 16 x > 43, 200 y>0 Can you tell why x > 0 is not an expression needed in the system of inequalities?

Exercise 9 Part (d) • Express the cost of buying x pounds of alfalfa hay and y pounds of corn (i. e. State the objective function for the problem) • Solution: Cost = 0. 01 x + 0. 016 y (Note: In the modified version of the steps, this is would occur after the definitions of the variables)

Exercise 9 Part (c) • Graph the feasible set for the system of linear inequalities. • Solution: 0. 13 x + 0. 065 y > 4, 550 0. 48 x + 0. 96 y > 26, 880 2. 16 x > 43, 200 y>0 y > – 2 x + 70, 000 y > – 1/2 x + 28, 000 x > 20, 000 y>0