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Linear Programming

Linear Programming. Macon State College Mary Dwyer Wolfe, Ph.D. July 2009. What is Linear Programming?. From Wikipedia :

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Linear Programming

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  1. Linear Programming Macon State College Mary Dwyer Wolfe, Ph.D. July 2009

  2. What is Linear Programming? From Wikipedia: In mathematics, linear programming (LP) is a technique for optimization of a linear objective function, subject to linear equality and linear inequality constraints. Informally, linear programming determines the way to achieve the best outcome (such as maximum profit or lowest cost) in a given mathematical model and given some list of requirements represented as linear equations.

  3. Why Teach Linear Programming? • Skills employed • Graphing equations and inequalities in 2 variables • Finding intercepts • Exploring slope and y-intercept • Standard form and slope-intercept form of lines • Finding solutions to systems of equations • Translating a problem from a verbal representation, to a symbolic representation and then to a graphical representation.

  4. Linear Programming Example Maximize I = 0.06x + 0.09y Subject to the constraints: x ≥ 0, y ≥ 0 x + y ≤ 25 x ≥ 15 y ≤ 5 x + y = 25 Intercepts are (0, 25) and (25, 0) (0, 0) make the inequality true

  5. Linear Programming Example Maximize I = 0.06x + 0.09y Subject to the constraints: x ≥ 0, y ≥ 0 x + y ≤ 25 x ≥ 15 y ≤ 5 x + y = 25 y = 5 x + 5 = 25 x = 20 If a Linear Programming Problem has a solution, it is located at a corner point of the feasible points.

  6. Linear Programming Example Maximize I = 0.06x + 0.09y If a Linear Programming Problem has a solution, it is located at a corner point of the feasible points. Use: (15, 5) (15, 0) (25, 0) (20,5)

  7. Linear Programming Example If a Linear Programming Problem has a solution, it is located at a corner point of the feasible points. Maximize I = 0.06x + 0.09y Use: (15, 5) (15, 0) (25, 0) (20,5) I = 0.06(15) + 0.09(5) = 1.35 I = 0.06(15) + 0.09(0) = 0.9 I = 0.06(25) + 0.09(0) = 1.5 I = 0.06(20) + 0.09(5) = 1.65 x = 20 and y = 5 maximizes I.

  8. Try this one! Maximize z = 3x + 5y Subject to: x ≥ 0, y ≥ 0, x + y ≥ 2, 2x + 3y ≤ 12, 3x + 2y ≤ 12

  9. Try this one! z is maximized when x = 0 and y = 4 Maximize z = 3x + 5y Subject to: x ≥ 0, y ≥ 0, x + y ≥ 2, 2x + 3y ≤ 12, 3x + 2y ≤ 12

  10. Using the Calculator While the calculator is not the best too for visualization in these problems, it can help find difficult intersection points.

  11. Now for a Real Problem Maximizing Profit At the end of every month, after filling orders for its regular customers, a coffee company has some pure Colombian coffee and some special-blend coffee remaining. The practice of the company has been to package a mixture of the two coffees into 1-pound packages as follows: a low-grade mixture containing 4 ounces of Colombian and 12 ounces of special-blend coffee and a high grade mixture containing 8 ounces of Colombian and 8 ounces of special-blend coffee. A profit of $0.30 per package is made on the low-grade mixture whereas a profit of $0.40 per package is made on the high grade mixture. This month, 120 pounds of special-blend coffee and 100 pounds of pure Columbian coffee remain. How many packages of each mixture should be prepared to achieve a maximum profit? Assume that all packages prepared can be sold.

  12. The Coffee Problem Solution Let x be the number of packages of low-grade mixture Let y be the number of packages of high-grade mixture Function to maximize is the Profit Function: P = 0.30x + 0.40y The profit is subject to the following constraints: x ≥ 0, y ≥ 0 Nonnegative constraints Total amount of Columbian coffee cannot exceed 100 pounds (1600 ounces). 4x + 8y ≤ 1600 Columbian coffee constraint Total amount of special blend coffee cannot exceed 120 pounds or 1920 ounces. 12x + 8y ≤ 1920 Special-blend coffee constraint

  13. The Coffee Problem Solution Let x be the number of packages of low-grade mixture Let y be the number of packages of high-grade mixture Maximize P = 0.30x + 0.40y Given the following constraints: x ≥ 0, y ≥ 0 4x + 8y ≤ 1600 12x + 8y ≤ 1920

  14. The Coffee Problem Solution Let x be the number of packages of low-grade mixture Let y be the number of packages of high-grade mixture Maximize P = 0.30x + 0.40y Given the following constraints: x ≥ 0, y ≥ 0 4x + 8y ≤ 1600 12x + 8y ≤ 1920 Profit can be maximized with 40 ounces of the low-grade mixture and 180 ounces of the high-grade mixture.

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