The following intermediate tableau for a dual maximum problem was obtained using the simplex method for optimizing a minimum problem. \[ \left[\begin{array}{cccccc}P & y_{1} & y_{2} & s_{1} & s_{2} & R H S \\ 0 & 1 & -\frac{4}{3} & \frac{1}{3} & 0 & \frac{2}{3} \\ 0 & 0 & 1 & -\frac{1}{10} & \frac{3}{10} & \frac{2}{5} \\ \hline 1 & 0 & -10 & 2 & 0 & 4\end{array}\right] \] Perform all the necessary pivot and row operations to obtain the final tableau. Then, using the final tableau, answer the following questions: The minimum function value is: The value of \( x_{1} \) in the minimum problem is : The value of \( x_{2} \) in the minimum problem is:

To solve the dual maximum problem using the given intermediate tableau, we will perform the necessary pivot and row operations to obtain the final tableau. The tableau is structured as follows: \[ \left[\begin{array}{cccccc} P & y_{1} & y_{2} & s_{1} & s_{2} & R H S \\ 0 & 1 & -\frac{4}{3} & \frac{1}{3} & 0 & \frac{2}{3} \\ 0 & 0 & 1 & -\frac{1}{10} & \frac{3}{10} & \frac{2}{5} \\ \hline 1 & 0 & -10 & 2 & 0 & 4 \end{array}\right] \] ### Step 1: Identify the pivot column The pivot column is determined by selecting the most negative coefficient in the objective function row (last row). Here, the most negative coefficient is \(-10\) corresponding to \(y_2\). ### Step 2: Identify the pivot row To find the pivot row, we calculate the ratio of the right-hand side (RHS) to the corresponding positive entries in the pivot column: - For the first row: \(\frac{\frac{2}{3}}{-\frac{4}{3}} = -\frac{1}{2}\) (not valid since it's negative) - For the second row: \(\frac{\frac{2}{5}}{1} = \frac{2}{5}\) The pivot row is the second row since it has the smallest non-negative ratio. ### Step 3: Perform the pivot operation The pivot element is \(1\) (the entry in the second row and second column). We will perform row operations to make all other entries in the pivot column \(0\). 1. **Row 1**: Replace \(R_1\) with \(R_1 - (-\frac{4}{3})R_2\): \[ R_1 = R_1 + \frac{4}{3}R_2 \] \[ R_1 = \left[0, 1, -\frac{4}{3}, \frac{1}{3}, 0, \frac{2}{3}\right] + \frac{4}{3}\left[0, 0, 1, -\frac{1}{10}, \frac{3}{10}, \frac{2}{5}\right] \] \[ R_1 = \left[0, 1, 0, \frac{1}{3} - \frac{4}{30}, \frac{0 + \frac{12}{30}}{1}, \frac{2}{3} + \frac{8}{15}\right] \] \[ R_1 = \left[0, 1, 0, \frac{1}{3} - \frac{2}{15}, \frac{12}{30}, \frac{10}{15}\right] = \left[0, 1, 0, \frac{1}{6}, \frac{2}{5}\right] \] 2. **Row 3**: Replace \(R_3\) with \(R_3 - 1R_2\): \[ R_3 = R_3 - R_2 \] \[ R_3 = \left[0, 0, 1, -\frac{1}{10}, \frac{3}{10}, \frac{2}{5}\right] - \left[0, 0, 1, -\frac{1}{10}, \frac{3}{10}, \frac{2}{5}\right] \] \[ R_3 = \left[0, 0, 0, 0, 0, 0\right] \] 3. **Objective Row**: Replace \(R_4\) with \(R_4 + 10R_2\): \[ R_4 = R_4 + 10R_2 \] \[ R_4 = \left[1, 0, -10, 2, 0, 4\right] + 10\left[0, 0, 1, -\frac{1}{10}, \frac{3}{10}, \frac{2}{5}\right] \] \[ R_4 = \left[1, 0, 0, 2 - 1, 0 + 3, 4 + 4\right] = \left[1, 0, 0, 1, 3, 8\right] \] ### Final Tableau After performing the necessary operations, the final tableau is: \[ \left[\begin{array}{cccccc} P & y_{1} & y_{2} & s_{1} & s_{2} & R H S \\ 0 & 1 & 0 & \frac{1}{6} & \frac{2}{5} & \frac{2}{3} \\ 0 & 0 & 1 & -\frac{1}{10} & \frac{3}{10} & \frac{2}{5} \\ \hline 1 & 0 & 0 & 1 & 3 & 8 \end{array}\right] \] ### Step 4: Extract the results From the final tableau, we can extract the values for the minimum function value and the values of \(x_1\) and \(x_2\): - The minimum function value is \(8\). - The value of \(x_1\) in the minimum problem is \(3\). - The value of \(x_2\) in the minimum problem is \(0\) (since \(y_2\) is not a basic variable). ### Final Answers - The minimum function value is: \(8\) - The value of \(x_{1}\) in the minimum problem is: \(3\) - The value of \(x_{2}\) in the minimum problem is: \(0\)