Chapter 3 Section 2 Linear Programming I

Fundamental Theorem of Linear Programming The maximum (or minimum) value of the objective function is achieved at one of the vertices

Exercise 21 (page 131) Minimize the objective function 3x + 4y subject to the below constraints. Solution: 2x + y > 10y > – 2x + 10 x + 2y > 14y > – ½ x + 7 x > 0 y > 0

Graph of the Inequalities y = – 2x + 10 y = – ½ x + 7 y = 0 x = 0 I II III

Finding the Vertices (by hand) Vertex IVertex IIVertex III x = 0 y = – 2x + 10y = – ½ x + 7 y = – 2x + 10y = – ½ x + 7y = 0 y = – 2(0) + 10 – 2x + 10 = – ½ x + 7( 0 ) = – ½ x + 7 y = 103/2 x = 3 ½ x = 7 x = 2 x = 14 ( 0, 10 ) y = – 2( 2 ) + 10 ( 14, 0 ) y = 6 ( 2, 6 )

Find the Optimal Point VertexObjective Function: 3 x + 4 y ( 0, 10 )3 ( 0 ) + 4 ( 10 ) = 40 ( 2, 6 )3 ( 2 ) + 4 ( 6 ) = 30 ( 14, 0 )3 ( 14 ) + 4 ( 0 ) = 42 The minimum value of the objective function occurs at the vertex ( 2, 6 )

Exercise 33 (page 132) Define the variables being used. Look at the question being asked! Last sentence: “How many cans of Fruit Delight and Heavenly Punch should be produced…?” Let x represent the number of cans of Fruit Delight produced y represent the number of cans of Heavenly Punch produced

Last sentence: “How many cans of Fruit Delight and Heavenly Punch should be produced each week to maximize profits?” The objective function is: Profit = 0.20 · x · y The Objective Function

Table Fruit Delight Heavenly Punch Maximum Amount Pineapple Juice Orange Juice Apricot Juice Profit

Table Fruit Delight Heavenly Punch Maximum Amount Pineapple Juice 10 oz/can Orange Juice 3 oz/can Apricot Juice 1 oz/can Profit0.20 $

Table Fruit Delight Heavenly Punch Maximum Amount Pineapple Juice 10 oz/can 10 oz/can Orange Juice 3 oz/can 2 oz/can Apricot Juice 1 oz/can 2 oz/can Profit0.20 $ 0.30 $

Table Fruit Delight Heavenly Punch Maximum Amount Pineapple Juice 10 oz/can 10 oz/can 9,000 oz Orange Juice 3 oz/can 2 oz/can 2,400 oz Apricot Juice 1 oz/can 2 oz/can 1,400 oz Profit0.20 $ 0.30 $

Restrictions from the Table and Problem Restrictions that are placed on the x and y variables: 10 x + 10 y < 9,000 3 x + 2 y < 2,400 x + 2 y < 1,400 x > 0 y > 0

Change From General Form to Standard Form 10 x + 10 y < 9,000 y < – x x + 2 y < 2,400 y < – 3/2 x + 1,200 x + 2 y < 1,400 y < – ½ x x > 0 y > 0

Graph of the System of Equations Not to scale y = – 3/2 x y = – x y = – 1/2 x y = 0 x = 0

Shading for the Inequalities Not to scale y = – 3/2 x y = – x y = – 1/2 x y = 0 x = 0

A Modified Graph of the Feasible Set from the Previous Slide y = 0 y = – 1/2 x y = – x y = – 3/2 x x = 0 I II III IV V Feasible Set

Finding the Vertices (using a calculator when possible) Vertex Ix = 0 ( 0, 0 ) y = 0 Vertex IIx = 0 ( 0, 700 ) y = – ½ x Vertex IIIy = – ½ x + 700( 400, 500 ) y = – x Vertex IVy = – x + 900( 600, 300 ) y = – 3/2 x Vertex Vy = – 3/2 x ( 800, 0 ) y = 0

Find the Optimal Point VertexObjective Function 0.2 x y ( 0, 0 )0.2 ( 0 ) ( 0 ) = 0 ( 0, 700 )0.2 ( 0 ) ( 700 ) = 210 ( 400, 500 )0.2 ( 400 ) ( 500 ) = 230 ( 600, 300 )0.2 ( 600 ) ( 300 ) = 210 ( 800, 0 )0.2 ( 800 ) ( 0 ) = 160 ( 400, 500 ) maximizes the objective function

Answer Produce 400 cans of Fruit Delight and 500 cans of Heavenly Punch to maximize profits

Exercise 35 (page 133) Define the variables being used. Look at the question being asked! Key Question: “…, what planting combination will produce the greatest total profit?” Let x represent the number of acres of oats planted y represent the number of acres of corn planted This now defines Column Headings

The Objective Function First we need to recognize these relationships Profit = Revenue + Left over capital cash + Left over labor costs Where –Revenue = 55 x y –Left over capital cash = 2100 – 18 x – 36 y –Left over labor cash = 2400 – 16 x – 48 y

The Objective Function Now add revenue, left over capital cash, and left over labor costs together [i.e. ( 55 x y ) + (2100 – 18 x – 36 y ) + ( 2400 – 16 x – 48 y )] to get the profit and the objective function becomes: Profit = x + 41 y

Table OatsCornAvailable Capital Needed ($) Labor Needed ($) Revenue ($)

Table OatsCornAvailable Capital Needed ($) 18 $/acre 36 $/acre 2,100 $ Labor Needed ($) 16 $/acre 48 $/acre 2,400 $ Revenue ($)55 $/acre 125 $/acre

Restrictions from Table and Problem The inequalities that restrict the values of the variables: 18 x + 36 y < 2, x + 48 y < 2,400 x + y < 100 x > 0 y > 0 Why do we need this restriction?

Why x + y < 100? The farmer has only 100 acres available for the two crops. The amount of oats plus the amount of corn that the farmer plants has to be 100 acres or less.

Convert from General Form to Standard Form Restrictions: 18 x + 36 y < 2,100 y < – ½ x + 175/3 16 x + 48 y < 2,400 y < – 1/3 x + 50 x + y < 100 y < – x x > 0 y > 0

Graph of the System of Equations Not to scale y = – x y = – ½ x + 175/3 y = – 1/3 x + 50 y = 0 x = 0

Graph of the System of Inequalities Not to scale y = – x y = – ½ x + 175/3 y = – 1/3 x + 50 y = 0 x = 0

Modified Graph of the Feasible Set form the Previous Slide y = 0 y = – 1/3 x + 50 y = – ½ x + 175/3 y = – x x = 0 I II III IV V Feasible Set

Finding the Vertices (using a calculator when possible) Vertex Ix = 0 ( 0, 0 ) y = 0 Vertex IIx = 0 ( 0, 50 ) y = – 1/3 x + 50 Vertex IIIy = – 1/3 x + 50( 50, 100/3 ) y = – ½ x + 175/3 Vertex IVy = – ½ x + 175/3( 250/3, 50/3) y = – x Vertex Vy = – x + 100( 100, 0 ) y = 0

Finding the Optimal Point VertexObjective Function x + 41y ( 0, 0 ) ( 0 ) + 41 ( 0 ) = 4, ( 0, 50 ) ( 0 ) + 41 ( 50 ) = 6, ( 50, 100/3 ) ( 50 ) + 41 ( 100/3 ) ~ 6, ( 250/3, 50/3 ) ( 250/3 ) + 41 ( 50/3 ) ~ 6, ( 100, 0 ) ( 100 ) + 41 ( 0 ) = 6, ( 250/3, 50/3 ) = ( 83 1/3, 16 2/3 ) maximizes the objective function

The Answer Plant 83-and-a-third acres of oats and 16- and-two-thirds acres of corn to maximize the profit