Use the simplex method to solve the following minimum problem on your own paper. Then, using your final tableau, enter the answer in each relevant box provided below. Minimize: \( C=x_{1}+2 x_{2} \) Subject to the following constraints: \( x_{1}+x_{2} \geq 6 \) \( 2 x_{1}+4 x_{2} \geq 12 \) \( x_{1} \geq 0 \quad ; \quad x_{2} \geq 0 \) Minimum value of \( C=6 \) Value of \( x_{1}=3 \) Value of \( x_{2}=3 \)

To solve the given linear programming problem using the simplex method, we first need to convert the problem into a standard form suitable for the simplex algorithm. ### Step 1: Convert the inequalities to equalities The constraints given are: 1. \( x_{1} + x_{2} \geq 6 \) 2. \( 2x_{1} + 4x_{2} \geq 12 \) We can convert these inequalities into equalities by introducing slack variables. However, since these are "greater than or equal to" constraints, we will introduce surplus variables instead. 1. For the first constraint, we introduce a surplus variable \( s_1 \): \[ x_{1} + x_{2} - s_1 = 6 \] 2. For the second constraint, we introduce a surplus variable \( s_2 \): \[ 2x_{1} + 4x_{2} - s_2 = 12 \] ### Step 2: Set up the objective function The objective function to minimize is: \[ C = x_{1} + 2x_{2} \] ### Step 3: Formulate the tableau We will rewrite the objective function in terms of the surplus variables: \[ C + 0s_1 + 0s_2 = 0 \] Now we can set up the initial simplex tableau: \[ \begin{array}{c|cccccc|c} \text{Basic Variables} & x_{1} & x_{2} & s_1 & s_2 & C & \text{RHS} \\ \hline s_1 & 1 & 1 & 1 & 0 & 0 & 6 \\ s_2 & 2 & 4 & 0 & 1 & 0 & 12 \\ \hline -C & -1 & -2 & 0 & 0 & 1 & 0 \\ \end{array} \] ### Step 4: Perform the simplex method 1. Identify the pivot column (most negative in the last row): here, it is \( x_{2} \) (column 2). 2. Calculate the ratios for the pivot row: - For \( s_1 \): \( \frac{6}{1} = 6 \) - For \( s_2 \): \( \frac{12}{4} = 3 \) The smallest ratio is 3, so we pivot on \( x_{2} \) in the second row. ### Step 5: Pivoting We will perform row operations to make the pivot column a unit column. 1. Divide the second row by 4: \[ \text{New } s_2: \quad \left( \frac{1}{2}, 1, 0, \frac{1}{4}, 0, 3 \right) \] 2. Update the first row: \[ \text{New } s_1: \quad \left( 1 - 0, 1 - 0, 1 - 0, 0 - 0, 0 - 0, 6 - 0 \right) = (1, 0, 1, -\frac{1}{4}, 0, 6) \] 3. Update the last row: \[ \text{New } -C: \quad \left( -1 + 2 \cdot \frac{1}{2}, -2 + 2 \cdot 1, 0, 0 + 0, 1, 0 + 3 \right) = (0, 0, 0, \frac{1}{2}, 1, 3) \] The new tableau is: \[ \begin{array}{c|cccccc|c} \text{Basic Variables} & x_{1} & x_{2} & s_1 & s_2 & C & \text{RHS} \\ \hline s_1 & 1 & 0 & 1 & -\frac{1}{4} & 0 & 6 \\ x_{2} & \frac{1}{2} & 1 & 0 & \frac{1}{4} & 0 & 3 \\ \hline -C & 0 & 0 & 0 & \frac{1}{2} & 1 & 3 \\ \end{array} \] ### Step 6: Final tableau The final tableau indicates that the optimal solution has been reached since there are no negative coefficients in the last row. ### Step 7: Read the solution From the final tableau: - The minimum value of \( C \) is \( 3 \). - The values of the variables are: - \( x_{1} = 3 \) - \( x_{2} = 3 \) ### Conclusion The final answers are: - Minimum value of \( C = 3 \) - Value of \( x_{1} = 3 \) - Value of \( x_{2} = 3 \) Please note that the provided values in the question were incorrect. The correct minimum value of \( C \) is \( 3 \), not \( 6 \).