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Posted by on Saturday, January 29, 2011 at 12:26am.

Minimize c = x + y subject to
x + 5y ≥ 6
5x + y ≥ 6
x ≥ 0, y ≥ 0.

  • math - , Saturday, January 29, 2011 at 9:32am

    graph the two lines
    x + 5y ≥ 6 (intercepts at (0,6/5)(6,0))
    5x + y ≥ 6 (intercepts at ((6/5,0)(0,6)
    x ≥ 0, y ≥ 0.

    find intersection
    5 x + 25 y = 30
    5 x + y = 6
    ------------------
    24 y = 24
    y = 1
    x = 6-5 = 1
    so at (1,1)
    evaluate c = x+y at three points
    (0,6) (1,1) (6,0)
    minimum is obviously 2 at (1,1)

  • linear programming - , Saturday, January 29, 2011 at 9:33am

    There is a name for that - "linear programming"

  • math - , Saturday, January 29, 2011 at 9:41am

    It is similar to the previous problems. You could plot them on a graph, find the intersection(s) and evaluate the objective function Z=x+y.

    The lines to plot are:
    for x+5y≥6, put the inequality in the standard form: y≥(6-x)/5.
    Similarly, for 5x+y≥6, plot it as y≥(6-5x).
    There are lines x≥0 and y≥0, which come into play whne x≥6 or y≥6.

    Plot the graph and find all possible intersections (there are three). The feasibility region is an open region in the first quadrant (towards the top-right).
    Now evaluate the objective function Z=x+y at all three intersections and find the point that gives the smallest value of Z.

    See:
    http://img9.imageshack.us/img9/7284/1296278783.png

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