D1,L9 Solving Linear Programming Problems

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    Decision Maths

    Solving Linear

    Programming problems

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    Solving the problem.

    There are a number of ways in which to solve the

    linear programming problem that we looked at in the

    last lesson.

    1 You can draw the equations graphically and theni Solve by evaluating the vertices.

    ii Solve using the ruler method.

    2 You can apply the Simplex Algorithm (this is not

    in the D1 syllabus and we will not be studying this) Taking last weeks example, we are going to look at

    representing the algebra graphically and then we will

    consider both methods to solve this problem.

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    Final - Problem

    Lets consider the example we looked at in lesson 7.

    Maximise the profit function.

    P = 60x + 84y

    Subject to the constraints

    2x + 3y 30

    5x + 5y 60

    x 0, y 0

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    To draw 5x + 5y = 60 and 2x + 3y = 30 find the intersections with both axes.

    i.e x=0 y=12 and y=0 x=12 the line joins (0,12) and (12,0).

    And x=0 y=10 and y=0 x=15 the line joins (0,10) and (15,0)

    The Graph

    (drawing the lines)

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    CraftsmanMachine

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    The Graph

    (shading inequalities)

    As we know that x and y are both 0, we can shade in the

    region of co-ordinates that we do not require.

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    The Graph

    (shading inequalities)

    We can now do a similar thing with the constraint inequalities.

    If2x + 3y 30, then choose any point and plug in to the inequality.

    Using the point (2,2), 2 x 2 + 3 x 2 = 10, which is less than 30. Fromthis we know to shade in the region above the line.

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    The Feasible Region

    We can do a similar thing with 5x + 5y 60, and

    again we shade in the region above the line.

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    Questions

    1 Indicate on a diagram the region for which

    5x + 3y 15

    x 0, y 0

    2 - Indicate on a diagram the region for which4x + 3y 12

    2x + 5y 10

    x 0, y 0

    3 - Indicate on a diagram the region for which2x + y 8

    y 7, x 3

    x 0, y 0

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    Questions

    4 Indicate on a diagram the region for which

    y + 2x 12

    x 2, y 4

    5 - Indicate on a diagram the region for which

    3x + 2y 12

    3x + y 6

    x + y 4

    6 - Indicate on a diagram the region for which

    3x + 2y 6

    y 2x

    y 0

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    The Feasible Region

    The feasible region is all of the co-ordinates that lie in the un-

    shaded area. Remember that each co-ordinate (x,y) represents

    the number of each shed made.

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    The Feasible Region

    The vertices of the feasible region are: (0,0), (12,0), (6,6) and (0,10).

    The co-ordinate (6,6) can be found by solving the simultaneous

    equations 2x + 3y = 30 and 5x + 5y = 60.

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    Simultaneous Equations

    (6,6) is found by solving 2x+3y=30 and 5x+5y=60simultaneously.

    2x + 3y = 30

    5x + 5y = 60

    2x + 3y = 30

    3x + 3y = 36

    x = 6

    2 x 6 + 3y = 30

    12 + 3y = 30

    3y = 18

    y = 6

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    Maximising the Profit

    FunctionMethod i

    The maximum profit will come from the values of x and y at oneof the vertices.

    We can now find the value of the profit function at each of thesepoints:

    the maximum profit of 864 can be obtained by making 6sheds of each type each day.

    x y P =60x+84y0 0 0

    12 0 720

    6 6 8640 10 840

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    Maximising the Profit

    FunctionMethod ii

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    The ruler Method

    You draw the object function 60x + 84y = P. (here pick P to be

    any value to help you, it makes no difference)

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    Maximising the Profit

    FunctionMethod ii

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    The ruler Method

    Once you have drawn the line move it to the last point available

    in the region keeping the line parallel.

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    Maximising the Profit

    FunctionMethod ii

    The co-ordinate that it lands on will give you

    the values of x and y that will optimise the

    function.

    Here, as we know already x = 6, y = 6. So P = 60x + 84y

    = 60 x 6 + 84 x 6

    = 864