Use the simplex method to solve the following maximum problem: Maximize: \( \quad P=4 x_{1}+9 x_{2} \) Subject to the constraints: \( \left\{\begin{array}{l}2 x_{1}+5 x_{2} \leq 9 \\ x_{1}+2 x_{2} \leq 4 \\ x_{1} \geq 0 \\ x_{2} \geq 0\end{array}\right. \) and using your final tableau answer the questions below by entering the correct answer in each blank box. \( x_{1}=\square \) \( P=\square \)

\[ \textbf{Step 1. Convert inequalities to equalities by introducing slack variables.} \] Introduce slack variables \( x_3 \) and \( x_4 \) such that: \[ \begin{aligned} 2x_1+5x_2+x_3 &= 9, \\ x_1+2x_2+x_4 &= 4, \\ x_1,x_2,x_3,x_4 &\ge 0. \end{aligned} \] The objective function to maximize is: \[ P = 4x_1+9x_2. \] \[ \textbf{Step 2. Set up the initial simplex tableau.} \] We form the tableau with basic variables \( x_3 \) and \( x_4 \): \[ \begin{array}{c|cccc|c} & x_1 & x_2 & x_3 & x_4 & \text{RHS} \\ \hline \text{Row 1 (for \( x_3 \))} & 2 & 5 & 1 & 0 & 9 \\ \text{Row 2 (for \( x_4 \))} & 1 & 2 & 0 & 1 & 4 \\ \hline \text{Row 0 (Objective)} & -4 & -9 & 0 & 0 & 0 \\ \end{array} \] \[ \textbf{Step 3. Identify the entering variable.} \] Examine the bottom row (Row 0). The most negative coefficient is \(-9\) corresponding to \( x_2 \). Therefore, \( x_2 \) enters the basis. \[ \textbf{Step 4. Determine the leaving variable using the minimum ratio test.} \] Calculate the ratios for rows where the coefficient of \( x_2 \) is positive: \[ \text{Row 1: } \frac{9}{5} = 1.8, \quad \text{Row 2: } \frac{4}{2} = 2. \] The smallest ratio is \( 1.8 \) from Row 1, so \( x_3 \) leaves and \( x_2 \) enters. \[ \textbf{Step 5. Perform the pivot operation (pivot on the element in Row 1, Column \( x_2 \), which is 5).} \] \[ \textbf{i) Divide Row 1 by 5:} \] \[ \begin{aligned} x_1:&\quad \frac{2}{5}, \\ x_2:&\quad 1, \\ x_3:&\quad \frac{1}{5}, \\ x_4:&\quad 0, \\ \text{RHS}:&\quad \frac{9}{5}. \end{aligned} \] Thus, the new Row 1 becomes: \[ \left(\frac{2}{5},\; 1,\; \frac{1}{5},\; 0,\; \frac{9}{5}\right). \] \[ \textbf{ii) Update Row 2 to eliminate \( x_2 \):} \] Subtract \( 2 \times \) (new Row 1) from Row 2: \[ \begin{array}{rcl} \text{Row 2: } (1,\, 2,\, 0,\, 1,\, 4) & - & 2\left(\frac{2}{5},\, 1,\, \frac{1}{5},\, 0,\, \frac{9}{5}\right) \\ & = & \left(1-\frac{4}{5},\, 2-2,\, 0-\frac{2}{5},\, 1-0,\, 4-\frac{18}{5}\right) \\ & = & \left(\frac{1}{5},\, 0,\, -\frac{2}{5},\, 1,\, \frac{2}{5}\right). \end{array} \] \[ \textbf{iii) Update the objective function Row 0 to eliminate \( x_2 \):} \] Add \( 9 \times \) (new Row 1) to Row 0: \[ \begin{array}{rcl} \text{Row 0: } (-4,\, -9,\, 0,\, 0,\, 0) & + & 9\left(\frac{2}{5},\, 1,\; \frac{1}{5},\, 0,\, \frac{9}{5}\right) \\ & = & \left(-4+\frac{18}{5},\, -9+9,\; 0+\frac{9}{5},\, 0+0,\, 0+\frac{81}{5}\right) \\ & = & \left(-\frac{2}{5},\, 0,\, \frac{9}{5},\, 0,\, \frac{81}{5}\right). \end{array} \] The updated tableau now is: \[ \begin{array}{c|cccc|c} & x_1 & x_2 & x_3 & x_4 & \text{RHS} \\ \hline \text{Row 1: } x_2 & \frac{2}{5} & 1 & \frac{1}{5} & 0 & \frac{9}{5} \\ \text{Row 2: } x_4 & \frac{1}{5} & 0 & -\frac{2}{5} & 1 & \frac{2}{5} \\ \hline \text{Row 0 (Objective)} & -\frac{2}{5} & 0 & \frac{9}{5} & 0 & \frac{81}{5} \\ \end{array} \] \[ \textbf{Step 6. Check for optimality.} \] The bottom row still has a negative coefficient (\(-\frac{2}{5}\) for \( x_1 \)). Thus, we select \( x_1 \) as the entering variable. \[ \textbf{Step 7. Determine the leaving variable using the ratio test.} \] Consider the rows where the coefficient of \( x_1 \) is positive: \[ \text{Row 1: } \frac{\frac{9}{5}}{\frac{2}{5}} = \frac{9}{2} = 4.5,\quad \text{Row 2: } \frac{\frac{2}{5}}{\frac{1}{5}} = 2. \] The smallest ratio is \( 2 \) from Row 2. Thus, \( x_4 \) leaves and \( x_1 \) enters. \[ \textbf{Step 8. Pivot on the element in Row 2, Column \( x_1 \) (which is \(\frac{1}{5}\)).} \] \[ \textbf{i) Divide Row 2 by \(\frac{1}{5}\):} \] Multiply Row 2 by 5: \[ \begin{aligned} x_1: &\quad 1, \\ x_2: &\quad 0, \\ x_3: &\quad -2, \\ x_4: &\quad 5, \\ \text{RHS}: &\quad 2. \end{aligned} \] Thus, new Row 2 becomes: \[ (1,\; 0,\; -2,\; 5,\; 2). \] \[ \textbf{ii) Update Row 1 to eliminate \( x_1 \):} \] Row 1 has a coefficient of \(\frac{2}{5}\) for \( x_1 \). Subtract \(\frac{2}{5}\) times the new Row 2 from Row 1: \[ \begin{array}{rcl} \text{Row 1: } \left(\frac{2}{5},\, 1,\, \frac{1}{5},\, 0,\, \frac{9}{5}\right) & - & \frac{2}{5}\times (1,\, 0,\,-2,\, 5,\, 2) \\ & = & \left(\frac{2}{5}-\frac{2}{5},\, 1-0,\, \frac{1}{5}+\frac{4}{5},\, 0-2,\; \frac{9}{5}-\frac{4}{5}\right) \\ & = & \left(0,\; 1,\; 1,\; -2,\; 1\right). \end{array} \] \[ \textbf{iii) Update the objective function Row 0 to eliminate \( x_1 \):} \] Row 0 has a coefficient of \(-\frac{2}{5}\) for \( x_1 \). Add \(\frac{2}{5}\) times the new Row 2 to Row 0: \[ \begin{array}{rcl} \text{Row 0: } \left(-\frac{2}{5},\, 0,\, \frac{9}{5},\, 0,\; \frac{81}{5}\right) & + & \frac{2}{5}\times (1,\, 0,\,-2,\, 5,\; 2) \\ & = & \left(-\frac{2}{5}+\frac{2}{5},\, 0+0,\, \frac{9}{5}-\frac{4}{5},\, 0+2,\; \frac{81}{5}+\frac{4}{5}\right) \\ & = & \left(0,\; 0,\; \frac{5}{5},\; 2,\; \frac{85}{5}\right) \\ & = & \left(0,\; 0,\; 1,\; 2,\; 17\right). \end{array} \] The updated tableau becomes: \[ \begin{array}{c|cccc|c} & x_1 & x_2 & x_3 & x_4 & \text{RHS} \\ \hline \text{Row 1: } x_2 & 0 & 1 & 1 & -2 & 1 \\ \text{Row 2: } x_1 & 1 & 0 & -2 & 5 & 2 \\ \hline \text{Row 0 (Objective)} & 0 & 0 & 1 & 2 & 17 \\ \end{array} \] \[ \textbf{Step 9. Read the solution from the final tableau.} \] The basic variables are: - \( x_2 \) (Row 1) with value \( 1 \), - \( x_1 \) (Row 2) with value \( 2 \). Nonbasic variables \( x_3 \) and \( x_4 \) are zero. Thus, we have: \[ x_1 = 2, \quad x_2 = 1, \quad P = 17. \] \[ x_{1} = 2 \quad \text{and} \quad P = 17. \]