Question: Solve the linear programming problem using the simplex method. Maximize P = 2x_1 + 3x_2 subject…

Solve the linear programming problem using the simplex method.

Maximize P = 2x_1 + 3x_2

subject to -2x_1 + x_2 ≤ 14 -x_1 + x_2 ≤ 35 x_2 ≤ 42 x_1, x_2 ≥ 0

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 of P is when x_1 = and x_2 = .

B. There is no optimal solution.

Solution

To solve the linear programming problem using the simplex method, the problem is as follows: Maximize: \[ P = 2x_1 + 3x_2 \] Subject to: \[ \begin{cases} -2x_1 + x_2 \leq 14 & \\ -x_1 + x_2 \leq 35 & \\ x_2 \leq 42 & \\ x_1, x_2 \geq 0 & \end{cases} \] Convert inequalities to equalities by adding slack variables: Introduce slack variables \(s_1\), \(s_2\), and \(s_3\): \[ \begin{cases} -2x_1 + x_2 + s_1 = 14 & \\ -x_1 + x_2 + s_2 = 35 & \\ x_2 + s_3 = 42 & \end{cases} \] Set up the initial simplex tableau: \[ \begin{array}{c|cccc|c} \text{Basis} & x_1 & x_2 & s_1 & s_2 & s_3 & \text{Solution} \\ \hline s_1 & -2 & 1 & 1 & 0 & 0 & 14 \\ s_2 & -1 & 1 & 0 & 1 & 0 & 35 \\ s_3 & 0 & 1 & 0 & 0 & 1 & 42 \\ \hline \text{P} & -2 & -3 & 0 & 0 & 0 & 0 \\ \end{array} \] Identify the pivot column: The most negative indicator in the bottom row is \(-3\) (under \(x_2\)), so column \(x_2\) is the pivot column. Determine the pivot row: Compute the ratio of the right-hand side to the pivot column: - Row 1: \( \frac{14}{1} = 14 \) - Row 2: \( \frac{35}{1} = 35 \) - Row 3: \( \frac{42}{1} = 42 \) The smallest ratio is 14, so row 1 is the pivot row. Perform row operations to make pivot column a unit vector: Pivot element is 1 (row 1, column \(x_2\)). Perform the row operations to make pivot column a unit vector: 1. Row 1: Divide by 1 2. Row 2: Add \((1) \times \text{new Row 1}\) to eliminate entry in pivot column 3. Row 3: No change needed as the pivot element is already zero New tableau after row operations: \[ \begin{array}{c|cccc|c} \text{Basis} & x_1 & x_2 & s_1 & s_2 & s_3 & \text{Solution} \\ \hline x_2 & -2 & 1 & 1 & 0 & 0 & 14 \\ s_2 & 1 & 0 & -1 & 1 & 0 & 21 \\ s_3 & 0 & 0 & 0 & 0 & 1 & 28 \\ \hline \text{P} & -8 & 0 & 3 & 0 & 0 & 42 \\ \end{array} \] Solution: The maximum value of \(P\) is 42 when \(x_1 = 0\) and \(x_2 = 14\).