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Linear Programming Vocabulary Linear Programming Objective Function Constraints

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Find the maximal and minimal value of P = 3x + 4y subject to the following constraints: The three inequalities in the curly braces are the constraints. The area of the plane that they mark off will be the feasibility region. The formula “P = 3x + 4y" is the optimization equation. I need to find the (x, y) corner points of the feasibility region that return the largest and smallest values of P.

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My first step is to solve each inequality for the more-easily graphed equivalent forms:

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It's easy to graph the system: graph the system

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To find the corner points -- which aren't always clear from the graph -- I'll pair the lines (thus forming a system of linear equations) and solve: system of linear equations y = –( 1 / 2 )x + 7 y = 3x y = –( 1 / 2 )x + 7 y = x – 2 y = 3x y = x – 2 –( 1 / 2 )x + 7 = 3x –x + 14 = 6x 14 = 7x 2 = x y = 3(2) = 6 –( 1 / 2 )x + 7 = x – 2 –x + 14 = 2x – 4 18 = 3x 6 = x y = (6) – 2 = 4 3x = x – 2 2x = –2 x = –1 y = 3(–1) = –3 Corner Point:(2,6) Corner Point (6,4) Corner Point: (-1,-3)

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So the corner points are (2, 6), (6, 4), and (–1, –3). Somebody really smart proved that, for linear systems like this, the maximum and minimum values of the optimization equation will always be on the corners of the feasibility region. So, to find the solution to this exercise, I only need to plug these three points into P= 3x + 4y

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(2, 6): z = 3(2) + 4(6) = 6 + 24 = 30 (6, 4): z = 3(6) + 4(4) = 18 + 16 = 34 (–1, –3): z = 3(–1) + 4(–3) = –3 – 12 = –15 Then the maximum of P= 34 occurs at (6, 4), and the minimum of P = –15 occurs at (–1, –3).

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Given the following constraints, maximize and minimize the value of P = –0.4x + 3.2y.

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First I'll solve the fourth and fifth constraints for easier graphing:

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The feasibility region looks like this:

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From the graph, I can see which lines cross to form the corners, so I know which lines to pair up in order to verify the coordinates. I'll start at the "top" of the shaded area and work my way clockwise around the edges: y = –x + 7 y = x + 5 y = –x + 7 x = 5 x = 5 y = 0 –x + 7 = x + 5 2 = 2x 1 = x y = (1) + 5 = 6 y = –(5) + 7 = 2[nothing to do] corner at (1, 6)corner at (5, 2)corner at (5, 0)

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30S Applied Math Mr. Knight – Killarney School Slide 1 Unit: Linear Programming Lesson 5: Problem Solving Problem Solving with Linear Programming Learning.

30S Applied Math Mr. Knight – Killarney School Slide 1 Unit: Linear Programming Lesson 5: Problem Solving Problem Solving with Linear Programming Learning.

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