Consider the following linear program max 01271 + 02x2, subject to 12:31 + 32:2 S 60, 31:1 _ $2 2 _71 5172 S 10; 11:1 2 0,322 free. The stande form LP is given by max 61311 + 62(312 ya), subject to 12311 + 3(y2 3,13) + 3J4 = 60, 3y1 (yz - ya) 1/5 = -7, (3/2 93) + 316 = 10, y 2 0. (a) Use row operations to show that 342 = 7 + 33/1 + 1/3 1/5 94 = 39 2191 + 32/5 96 = 3 33/1 + 315 [Hint: You will notice that the columns corruponding to variables 3/4 and 3/6 are already columns of the identity matrix. So, all you have to do is convert the convert the column corresponding to yg into the remaining column of identity.] From this set of equations it follows that B = {y2,y4, ya} is a basis. Using only the information in the above set of equalities, and no matrix inversions compute the reduced cost with respect to each of the non-basic directions N = {311, 313,395} in terms of the (01,02). Recall that the reduced cost is the impact on the objective when any one of the non-basic variaqu is increased. (b) Describe the set of (c1, Cg) for which the basis B = {y2, y4, y6} is optimal. (0) Fix one c for which the current basis is optimal. Suppose the RHS 60 1 b3 2 7 + 9 2 10 1 What is the range of 6 for which this basis remains Optimal? Do not use matrix inversions to establish the result instead use the same row Operations that you used in part (a)