1. **State the problem:** We want to maximize the objective function $$Z = x_1 + 9x_2 + x_3$$ subject to the constraints: $$\begin{cases} x_1 + 2x_2 + 3x_3 \leq 9 \\ 3x_1 + 2x_2 + 2x_3 \leq 15 \\ 2x_1 + 3x_2 + x_3 \leq 14 \\ x_1, x_2 \geq 0 \end{cases}$$ 2. **Formulate the problem for the Revised Simplex Method:** Introduce slack variables $s_1, s_2, s_3 \geq 0$ to convert inequalities to equalities: $$\begin{cases} x_1 + 2x_2 + 3x_3 + s_1 = 9 \\ 3x_1 + 2x_2 + 2x_3 + s_2 = 15 \\ 2x_1 + 3x_2 + x_3 + s_3 = 14 \end{cases}$$ 3. **Initial basic feasible solution:** Set $x_1 = x_2 = x_3 = 0$, then $s_1 = 9$, $s_2 = 15$, $s_3 = 14$. 4. **Set up matrices:** - Coefficient matrix $A$ for variables $x_1, x_2, x_3$: $$A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 2 & 2 \\ 2 & 3 & 1 \end{bmatrix}$$ - Slack matrix $I$ (identity for slack variables): $$I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ - Initial basis $B = I$, non-basic variables $N = A$. 5. **Objective function coefficients:** $$c = \begin{bmatrix} 1 & 9 & 1 \end{bmatrix}$$ Slack variables have zero coefficients. 6. **Iterate using Revised Simplex Method:** - Compute $B^{-1}$ (initially identity). - Compute reduced costs $\bar{c}_N = c_N - c_B B^{-1} N$. - Select entering variable with most positive reduced cost. - Compute direction vector $d = B^{-1} a_j$. - Compute step length $\theta = \min \{ x_B / d_i | d_i > 0 \}$. - Update basis by replacing leaving variable. 7. **Perform calculations:** - Initial reduced costs are $c_N$ since $c_B=0$. - $c_N = [1,9,1]$, so $x_2$ enters (largest coefficient 9). - Compute $d = B^{-1} a_2 = a_2 = [2,2,3]^T$. - Compute $\theta = \min \{9/2, 15/2, 14/3\} = \min \{4.5, 7.5, 4.666...\} = 4.5$. - Leaving variable is $s_1$. 8. **Update basis:** Replace $s_1$ with $x_2$. 9. **Repeat iterations:** - New basis matrix and inverse updated. - Compute new reduced costs. - Continue until no positive reduced costs remain. 10. **Final solution:** After performing the Revised Simplex iterations (detailed matrix calculations omitted for brevity), the optimal solution is: $$x_1 = 0, \quad x_2 = 4.5, \quad x_3 = 0, \quad Z = 1 \times 0 + 9 \times 4.5 + 1 \times 0 = 40.5$$ **Answer:** The maximum value of $Z$ is **40.5** at $x_1=0$, $x_2=4.5$, $x_3=0$.