MAX X Z = 3X1 + 4x2
s.t 2x1 + 2x2 ≤ 8
1x1 + 2x2 ≤ 6
2x2 ≥ 1
please graph with the optimal solution. Then dual price for the first constraint by adding one. Then dual price for the third constraint adding one. Please also graph these with the same graph showing the new optimal solutions Also please show the iso-z lines for the initial problem.
Graphical representation of the problem is following:
X1 is plotted along the x-axis and X2 along the y-axis
Feasible region is the dark shaded region bounded by corner points as highlighted on the graph.
Iso-profit lines are three parallel lines shown in red, blue and green color.
Optimal solution is: X1 = 2, X2 = 2 (it gives the maximum value of objective function of all corner points of the feasible region)
We see that, increasing the RHS of constraint 1 by 1 unit, results in increase in objective value by 1 unit (from 14 to 15)
Therefore, dual price of the first constraint = 1
This is shown in the following graph
We see that, increasing the RHS of constraint 3 by 1 unit, results in increase in objective value by 0 unit (from 14 to 14)
Therefore, dual price of the third constraint = 0
This is shown in the following graph:
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