1. **State the problem:** We need to minimize the cost function $$\text{Cost} = 1.80S + 2.20T$$ subject to the constraints: $$5S + 8T \geq 200$$ $$15S + 6T \geq 240$$ $$4S + 12T \geq 180$$ $$T \geq 10$$ 2. **Convert inequalities to standard form for the simplex method:** Since all constraints are \(\geq\), we introduce surplus and artificial variables to convert them into equalities: $$5S + 8T - S_1 + A_1 = 200$$ $$15S + 6T - S_2 + A_2 = 240$$ $$4S + 12T - S_3 + A_3 = 180$$ $$T - S_4 = 10$$ where \(S_i\) are surplus variables and \(A_i\) are artificial variables. 3. **Set up the initial simplex tableau and perform Phase 1 to remove artificial variables.** 4. **After performing simplex iterations, Table 3 corresponds to the third tableau in the process.** 5. **Identify the outgoing variable in Table 3:** The outgoing variable is the basic variable that leaves the basis during the pivot operation in Table 3. This is determined by the minimum ratio test on the pivot column. 6. **Answer:** The outgoing variable in Table 3 is \(S_2\). This means that in the third tableau, the surplus variable \(S_2\) leaves the basis to be replaced by the entering variable (which would be determined by the pivot column).