5X1 + 4X2 = 300 X1 >= 100 X., X >=0 > Solve the above problem with the simplex method. Provide the value of the basic variables(BV) at tableau #3 and the optimal solution of the LP. Notes: Set M = 100 Tableau #1 is the initial tableau where an inconsistency exists on row Z. Tableau #2 is the tableau where the inconsistency on row z is fixed. Set si as the stack variable for constraint 1. Select one: OBV: s1 1500, A1-200, x1 = 100 1 Optimal sol x1 = 300, X20. z = 7200 OBV: s1 - 2000, A1 = 300, A2 = 100 Optimal sol: x1 = 150, X2 = 0,2= 3600 None of the options provided BV:51=1500, A1 =150, x = 150 Optimal sol: x1 = 150, X2 = 0,2 = 3600 BV:51-500, 53200, x1100 Optimal sol x1 = 300, X2=0,2 = 7200 Clear my.choice 3:05 PM