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  1. #1
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    linear programming problem

    Consider the linear programming problem below:

    Max 3A + 2B
    s.t.
    1A + 1B is less than or equal to 10
    3A + 1B is less than or equal to 24
    1A + 2B is less than or equal to 16
    A, B is greater than or equal to 0

    a. Use Management Scientist to solve the problem.

    b. If the objective function coefficient for A changes from 3 to 5, does the optimal solution changes? why?

    c. If the objective function coefficient for A remains 3, but the objective function coefficient for B varies from 2 to 4, does the optimal solution changes? why?

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  2. #2
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    Quote Originally Posted by lillian View Post
    Consider the linear programming problem below:
    Max 3A + 2B
    s.t.
    1A + 1B is less than or equal to 10
    3A + 1B is less than or equal to 24
    1A + 2B is less than or equal to 16
    A, B is greater than or equal to 0

    a. Use Management Scientist to solve the problem.
    b. If the objective function coefficient for A changes from 3 to 5, does the optimal solution changes? why?
    c. If the objective function coefficient for A remains 3, but the objective function coefficient for B varies from 2 to 4, does the optimal solution changes? why?
    Hello,

    to a). I don't know a Management Scientist but I can give you a solution.

    I use x for your A and I use y for your B.

    1A + 1B is less than or equal to 10: x+y\leq 10~\Longrightarrow \boxed{y\leq -x+10}
    3A + 1B is less than or equal to 24: 3x+y\leq 24~\Longrightarrow \boxed{y\leq -3x+24}
    1A + 2B is less than or equal to 16: x+2y\leq 16~\Longrightarrow \boxed{y\leq -\frac{1}{2}x+8}
    A, B is greater than or equal to 0: x\geq 0~\wedge~y\geq 0

    I sketched the 3 straight lines which are the borders of the half-planes. You get an irregular pentagon (hmm... don't forget we are talking about math). The graph of the objective function must pass at least through one point of this pentagon.

    Max = 3A + 2B: y = -\frac{3}{2}x+\frac{Max}{2}. That means the y-intercept is \frac{Max}{2}. The larger the y-intercept the larger is Max. If the graph of the objective function passes through B(7, 3) then the y-intercept will be the largest.
    Calculate the equation of the obj. function (sketched in red):
    B(7,3) \in y = -\frac{3}{2}x+c~\Longrightarrow~ 3= -\frac{3}{2}\cdot 7+c ~\Longrightarrow c = 11.5. Therefore \frac{Max}{2} = 11.5~\Longrightarrow~Max = 23.

    to b).
    Max = 5A + 2B: y = -\frac{5}{2}x+\frac{Max}{2}. That means the y-intercept is \frac{Max}{2}. The larger the y-intercept the larger is Max. If the graph of the objective function passes through B(7, 3) then the y-intercept will be the largest.
    Calculate the equation of the obj. function (sketched in green):
    B(7,3) \in y = -\frac{5}{2}x+c~\Longrightarrow~ 3= -\frac{5}{2}\cdot 7+c ~\Longrightarrow c = 20.5. Therefore \frac{Max}{2} = 20.5~\Longrightarrow~Max = 41.

    to c).
    Max = 3A + 4B: y = -\frac{3}{4}x+\frac{Max}{4}. That means the y-intercept is \frac{Max}{4}. The larger the y-intercept the larger is Max. If the graph of the objective function passes through A(4, 3) then the y-intercept will be the largest.
    Calculate the equation of the obj. function (sketched in blue):
    B(4, 6) \in y = -\frac{3}{4}x+c~\Longrightarrow~ 6= -\frac{3}{4}\cdot 4+c ~\Longrightarrow c = 9. Therefore \frac{Max}{4} = 9~\Longrightarrow~Max = 36.
    Attached Thumbnails Attached Thumbnails linear programming problem-linopt_lillian.gif  
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