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Math Help - Linear Programming Problem!!

  1. #1
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    Linear Programming Problem!!

    Consider the following linear programming problem:

    Maximise Z = 100x1 + 200x2

    Subject to 3x1 + 2x2 ≤ 120
    x1 + 4x2 ≤ 80
    2x1 + x2 ≤ 100
    x1, x2 ≥ 0

    Solution is Z=5600
    x1=32
    x2=12
    s3=24

    If a new constraint 2x1+ 2x2 ≤ 100 was included in the original problem, how would this affect the solution?

    My answer is that there is no change. But is this correct and is tehre a quicker way of getting the answer rather than the long drawn-out simplex method??


    ***EDIT I have another question where there are 3 variables X,Y, and Z. How do I plot that/answer the question then??? ***
    Last edited by DooBeeDoo; March 20th 2008 at 12:23 PM.
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  2. #2
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    Hello, DooBeeDoo!

    Consider the following linear programming problem:

    Maximise: . Z \:= \:100x + 200y

    Subject to: . \begin{Bmatrix}3x + 2y & \leq &120 & [1]\\ x + 4y & \leq & 80 & [2]\\ 2x + y & \leq & 100 & [3]\\ x, y & \geq & 0 & [4]\end{Bmatrix}

    Solution is: . Z=5600,\;x =32,\;y=12,\;s3=24

    If a new constraint 2x+ 2y \leq 100 was included in the original problem,
    how would this affect the solution?

    My answer is that there is no change. . . . . I agree

    Is there a quicker way of getting the answer?
    . . You weren't shown how to graph these problems?

    Constraint [4] puts us in Quandrant 1.

    For [1]\;3x + 2y \:\leq \:120, graph the line: . 3x + 2y \:=\:120
    It has intercepts: (40, 0), (0, 60)
    Draw the line and shade the region below it.

    For [2]\;x + 4y \:\leq \:80 graph the line: . x + 4y \:=\:80
    It has intercepts: (80, 0), (0, 20)
    Draw the line and shade the region below it.

    The shaded region looks like this . . .
    Code:
            |
         60 *
            |*
            | *
            |  *
            |   *
            |    *
         20 o     *
            |:::*  *
            |:::::::o
            |::::::::*  *
            |:::::::::*     *
            |::::::::::*        *
        - - o - - - - - o - - - - - * - -
            |          40          80

    Constraint [3] is useless.
    The line 2x + y \:=\:100 has intercepts (50, 0), (0, 100)
    . . and falls outside of the shaded region.


    We are concerned with the vertices of the shaded region.
    We can see three of them: .(0, 0), (40, 0), (0, 20)

    We find the fourth by locating the intersection of the two slanted lines.

    We solve the system: . 32,\:12)" alt="\begin{array}{ccc}3x + 2y &=&120 \\ x + 4y &=& 80\end{array}\;\text{ and get: }\32,\:12)" />


    We substitute these coordinates into the Z-function
    . . to determine which one maximizes Z.

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  3. #3
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    thanks a lot dude. very helpful.
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  4. #4
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    I have another question where there are 3 variables X,Y, and Z. How do I plot that/answer the question then???
    Last edited by DooBeeDoo; March 20th 2008 at 12:22 PM.
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  5. #5
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    Hello, DooBeeDoo!

    I have another question where there are 3 variables X,Y, and Z.
    How do I plot that / answer the question then?

    With three or more variables, there are mechanical procedures.
    . . The most common is the Simplex Method.
    Do a Google search.

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