1. (25 points) Consider the following LP: max 2 = 3x1 + 2x2 (1) s.t. 2x1 + x2 <100 (2) 1280 (3) 140 21, 20 (4) (5) In order to bring the problem into standard form, we have to add the slack variables $1, $2 and $3. The optimal tableau is given in the following table: ] I2 $1 $2 $3 RHS Basic Variable 0 0 1 1 0 180 z = 180 0 1 0 1 -1 0 20 1=20 0 0 1 -1 2 0 60 x2 = 60 0 0 0 -1 1 1 20 83 = 20 (a) Show that as long as the objective function coefficient of variable is between 2 and 4, the current basis remains optimal. Find the optimal solution if the objective function coefficient of variable has value 3.5. (b) Show that as long as the objective function coefficient of variable 2 is between 1.5 and 3, the current basis remains optimal. (c) Show that if the RHS of the first constraint varies between 80 and 120, then the current basis remains optimal. Find the new optimal solution if the RHS of the first constraint changes to 90.