2. Consider the following primal LP: max z = - 4x1 X2 s.t; 4x1 + 3x2 2 6 X1 + 2x2 0 After subtracting an excess variable efrom the first constraint, adding a slack variable s to the second constraint, and adding artificial variables a and az to the first and third constraints, the optimal tableau for this primal LP is as shown below. Z Rhs X1 0 S2 1/5 X2 0 1 0 0 1 0 0 0 ei 0 0 0 1 a1 M 0 0 3/5 a2 M-7/5 -1/5 2/5 1 -18/5 6/5 3/5 0 1 0 -1/5 1 -1 b. In the primal LP, find the range for b3 (RHS value of the third constraint, currently bz=3) for which the current basis remains optimal. What is the shadow price for this constraint? What would the new z-value be if b3 were 8