Please do #2

Question 1 8 pts Here is the optimal tableau for a Max LP 2 22 21 0 23 0 $1 0 82 10 $3 10 1 5 rhs 280 24 0 0 - 2 0 1 2 -8 0 0 2 1 0 2 4 8 0 1 1.25 0 0 -.5 1.5 2 1. Using B-matrix method, the original rhs vector bhad these coordinate values: b1- : 62 ; and b3 2. If we increase the original c2 by 8 and update this (formerly) optimal tableau, the new rowo value for x2 is and so this tableau is not optimal using regular simplex, the new optimal values are x1 - : x2 and x3 3. After increasing the original value of b2 by two more than its allowable increase, we find that we have the following (B-matrix method) updated tableau: z 21 0 22 5 23 0 1 0 0 2 0 81 82 83 rhs 0 10 10 340 1 2 - 8 36 0 2 - 4 20 0 -1/2 3/2 - 1 0 0 - 2 1 5/4 0 0 1 we now have to use Dual Simplex (that should have been a question) to get the new optimal tableau with new optimal value for z