Essential Understanding and Objectives● Essential Understanding: Some real-world problems involve multiple linear relationships. Linear programming accounts for all of these linear relationships and gives the solution to the problem.● Objectives: ● Students will be able to: ● Solving problems using linear programming ● Define constraint, linear programming, feasible region, and objective function
Iowa Core Curriculum• Algebra• A.CED.3 Represent constraints by equations or inequalities, and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different foods.
Linear Programming Businesses use linearprogramming to find out how tomaximize profit or minimizecosts. Most have constraints onwhat they can use or buy.
Linear ProgrammingLinear programming is a process of findinga maximum or minimum of a function byusing coordinates of the polygon formedby the graph of the constraints.
What is a constraint? A restriction... A boundary… A limitation…
What is the feasible region?The feasible region is the area of thegraph in which all the constraints are met.
Objective Function• The quantity you are trying to maximize or minimize is modeled by this.• Usually this quantity is the cost or profit• Looks something like this C = ax + by, where a and b are real numbers
Vertex Principle • If there is a maximum or minimum value of the linear objective function, it occurs at one or more vertices of the feasible region. Online Example
Find the minimum and maximumvalue of the function f(x, y) = 3x - 2y.We are given the constraints:•y ≥ 2• 1 ≤ x ≤5•y ≤ x + 3
Linear Programming• Find the minimum and maximum values by graphing the inequalities and finding the vertices of the polygon formed.• Substitute the vertices into the function and find the largest and smallest values.
Linear Programming •f(0, 2) = 6 minimum •f(4, 3) = 25 maximum
Example 1A farmer has 25 days to plant cotton and soybeans.The cotton can be planted at a rate of 9 acres per day,and the soybeans can be planted at a rate of 12 acresa day. The farmer has 275 acres available. If theprofit for soybeans is $18 per acre and the profit forcotton is $25 per acre, how many acres of each cropshould be planted to maximize profits?
Step 1: Define the variables What are the unknown values?Let c = number of acres of cotton toplantLet s = number of acres of soybeans toplant
Step 2: Write a System ofInequalitiesWrite the constraints. What are thelimitations given in the problem? The number of acres planted in cotton must c 0 be greater than or equal to 0. s 0 The number of acres planted in soybeans must be greater than or equal to 0. c s 275 The total number of acres planted must be less than or equal to 275. c s 25 The time available for planting must be less 9 12 than or equal to 25 days.
Step 3: Graph theInequalities s c The purple area is the feasible region.
Step 4: Name the Vertices of the Feasible Region of the vertices of theFind the coordinatesfeasible region, the area inside the polygon. (0,275) (225,0) (0,0) (75,200)
Step 5: Write an Equation tobe Maximized orMinimized p(c,s) = 25c + 18s Maximum profit = $25 times the number of acres of cotton planted + $18 times the number of acres of soybeans planted.
Step 6: Substitute theCoordinates into theEquation the coordinates of the vertices into the Substitute maximum profit equation. (c,s) 25c + 18s f(c,s) (0,275) 25(0) + 18(275) 4950 (225,0) 25(225) + 18(0) 5625 (0,0) 25(0) + 18(0) 0 (75,200) 25(75) + 18(200) 5475
Step 7: Find the Maximum (c,s) 25c + 18s f(c,s) (0,275) 25(0) + 18(275) 4950 (225,0) 25(225) + 18(0) 5625 (0,0) 25(0) + 18(0) 0 (75,200) 25(75) + 18(200) 5475 225 acres of cotton and 0 acres of soybeans should be planted for a maximum profit of $5,625.
Example 2:The Bethlehem Steel Mill can convert steel intogirders and rods. The mill can produce at most100 units of steel a day. At least 20 girders andat least 60 rods are required daily by regularcustomers. If the profit on a girder is $8 and theprofit on a rod is $6, how many units of eachtype of steel should the mill produce each day tomaximize the profits?
Step 1: Define the Variables Let x = number of girders Let y = number of rodsStep 2: Write a System of Inequalities x 20 At least 20 girders are required daily. y 60 At least 60 rods are required daily. The mill can produce at most 100 units x y 100 of steel a day.
Step 3: Graph the Inequalitiesx 20, y 60, and x y 100 100 80 60 40 20The purple region represents the feasible region.-50 50 100
Step 4: Name the Vertices of theFeasible Region Find the coordinates of the vertices of the feasible region, the area inside the polygon. (20, 60) (20, 80) (40, 60)
Step 5: Write an Equation to be Maximized or Minimized p(x,y) = 8x + 6yMaximum profit = $8 times the number ofgirders produced + $6 times the number ofrods produced
Step 6: Substitute the Coordinates into the EquationSubstitute the coordinates of the verticesinto the maximum profit equation. (x,y) 8x + 6y p(x,y) (20, 60) 8(20) + 6(60) 520 (20, 80) 8(20) + 6(80) 640 (40, 60) 8(40) + 6(60) 680
Step 7: Find the Maximum (x,y) 8x + 6y p(x,y) (20, 60) 8(20) + 6(60) 520 (20, 80) 8(20) + 6(80) 640 (40, 60) 8(40) + 6(60) 680 40 girders and 60 rods of steel should be produced for a maximum profit of $680.