Question: (B) Write the simplex tableau by filling in the blanks below. \[ \begin{array}{c|ccccc|c} & x_1…

(B) Write the simplex tableau by filling in the blanks below.

\[ \begin{array}{c|ccccc|c} & x_1 & x_2 & s_1 & s_2 & P \\ \hline s_1 & 4 & 1 & 1 & 0 & 0 & 8 \\ s_2 & 1 & 5 & 0 & 1 & 0 & 8 \\ \hline P & -16 & -1 & 0 & 0 & 1 & 0 \end{array} \]

The pivot element is located in the \( x_1 \) column and \( s_1 \) row.

The entering variable is \( x_1 \). The exiting variable is \( s_1 \).

(C) Use the simplex method to solve the problem. Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

A. The maximum value is \( P = \boxed{\phantom{0}} \) when \( x_1 = \boxed{\phantom{0}} \) and \( x_2 = \boxed{\phantom{0}} \).

(Type integers or simplified fractions.)

B. There is no optimal solution.

Solution

To solve the problem using the simplex tableau, we need to follow the simplex method steps. First, identify the pivot element. The pivot is in the \(x_1\) column and \(s_1\) row. The element is 4. Perform row operations to get a 1 in the pivot location and zeros in the other entries of the pivot column. 1. Divide the first row by 4 to make the pivot element 1: \[ \begin{bmatrix} 1 & 0.25 & 0.25 & 0 & 0 & 2 \\ 5 & 0 & 1 & 0 & 1 & 8 \\ -16 & -1 & 0 & 1 & 0 & 0 \end{bmatrix} \] 2. Update the second and third rows to make the rest of the \(x_1\) column zeros. - Second row: Subtract \(5\) times the first row from the second row: \[ \begin{bmatrix} 1 & 0.25 & 0.25 & 0 & 0 & 2 \\ 0 & -1.25 & -0.25 & 0 & 1 & -2 \\ -16 & -1 & 0 & 1 & 0 & 0 \end{bmatrix} \] - Third row: Add \(16\) times the first row to the third row: \[ \begin{bmatrix} 1 & 0.25 & 0.25 & 0 & 0 & 2 \\ 0 & -1.25 & -0.25 & 0 & 1 & -2 \\ 0 & 3 & 4 & 1 & 0 & 32 \end{bmatrix} \] The new tableau shows that the basic feasible solution has changed. Now we need to perform feasibility checks and update variables as necessary. Check the objective row (third row) if there are any negative coefficients that allow us to increase the objective function. The coefficient for \(x_2\) is positive, meaning we can perform another pivot operation, entering \(x_2\). Continue with further row operations using the simplex method to determine final optimal values. With the current tableau, further calculations will be needed to reach the optimal solution. Based on the problem setup and the tableau, you may continue simplifying and choosing pivots as required by the simplex method. When all coefficients in the last row are non-negative, the tableau represents the optimal solution.