1. **Problem Statement:** We want to maximize the objective function $$w = 3x_1 + x_2$$ subject to constraints: $$x_1 \leq 5, \quad x_2 \leq 8, \quad x_i \leq 4, \quad x_i \geq 0, x_2 \geq 0$$ 2. **Starting Tableau Vector q:** The vector \(q\) corresponds to the coefficients of the objective function in the simplex tableau. Since the objective is $$w = 3x_1 + x_2$$, the vector \(q\) includes these coefficients for the decision variables and zeros for slack variables. Thus, \(q = (3,1,0)\). 3. **Starting Tableau r-row:** The \(r\)-row in the simplex tableau is the negative of the objective function coefficients for the decision variables, followed by zeros for slack variables. So, \(r = (-3,-1,0,0,0)\). 4. **Variable Entering and Leaving the Basis:** If \(x_2\) enters the basis, the leaving variable is determined by the minimum ratio test on the constraints. Given constraints and slack variables \(y_1, y_2, y_a\), the leaving variable is \(y_2\). 5. **Reasonableness of Inserting \(y_1\) into the Basis:** Since \(y_1\) is a slack variable and already basic, it is not reasonable to insert it again. Therefore, the answer is: no, because \(y_1\) is already a basic variable. 6. **Final Tableau Objective Function:** Using the simplex method, the final objective function expressed in non-basic variables is: $$w = 13 - 1y_2 - 2y_a$$ **Summary of answers:** (a) \(q = (3,1,0)\) (b) \(r = (-3,-1,0,0,0)\) (c) Leaving variable: \(y_2\) (d) No, because \(y_1\) is already a basic variable (e) \(w = 13 - 1y_2 - 2y_a\)