Using simplex method

Max z = x1+2x2

Subject to-x1+2x2<8 x1+2x2<12 x1-x2<3 x1, x2≥0

To solve the linear programming problem using the **Simplex Method**, we'll first reformulate the problem by converting the inequalities into equations using slack variables. Let's proceed step by step.

### Problem Statement:

**Maximize:**  
\[ Z = x_1 + 2x_2 \]

**Subject to:**  
\[
\begin{align*}
-x_1 + 2x_2 & \leq 8 \\
x_1 + 2x_2 & \leq 12 \\
x_1 - x_2 & \leq 3 \\
x_1, x_2 & \geq 0
\end{align*}
\]

### Step 1: Convert inequalities to equalities by introducing slack variables

Let \( s_1, s_2, s_3 \) be the slack variables for each inequality:

\[
\begin{align*}
-x_1 + 2x_2 + s_1 &= 8 \\
x_1 + 2x_2 + s_2 &= 12 \\
x_1 - x_2 + s_3 &= 3 \\
x_1, x_2, s_1, s_2, s_3 &\geq 0
\end{align*}
\]

### Step 2: Set up the initial Simplex tableau

We express the objective function in standard form:  
\[
Z - x_1 - 2x_2 = 0 \quad \text{(which becomes)} \quad Z + (-x_1 - 2x_2) = 0
\]

The initial Simplex tableau looks like this:

| Basic Variables | \(x_1\) | \(x_2\) | \(s_1\) | \(s_2\) | \(s_3\) | RHS |
|-----------------|--------|--------|--------|--------|--------|------|
| \(s_1\)         | -1     | 2      | 1      | 0      | 0      | 8    |
| \(s_2\)         | 1      | 2      | 0      | 1      | 0      | 12   |
| \(s_3\)         | 1      | -1     | 0      | 0      | 1      | 3    |
| \(Z\)           | -1     | -2     | 0      | 0      | 0      | 0    |

### Step 3: Perform the Simplex Method

1. **Entering variable**: Identify the most negative coefficient in the objective row. Here, \(x_2 = -2\) is the most negative, so \(x_2\) enters the basis.
2. **Leaving variable**: To determine which variable leaves the basis, compute the ratio of the RHS to the corresponding coefficient in the column for \(x_2\):
   - Row 1: \( \frac{8}{2} = 4 \)
   - Row 2: \( \frac{12}{2} = 6 \)
   - Row 3: \( \frac{3}{-1} = \) Not considered (negative)

   So, \(s_1\) leaves the basis (smallest ratio is 4).

3. **Pivot** on the element at row 1, column \(x_2\) (which is 2).

Let's update the tableau after performing the pivot operation on this element:

#### Pivoted Tableau:

| Basic Variables | \(x_1\) | \(x_2\) | \(s_1\) | \(s_2\) | \(s_3\) | RHS |
|-----------------|--------|--------|--------|--------|--------|------|
| \(x_2\)         | -0.5   | 1      | 0.5    | 0      | 0      | 4    |
| \(s_2\)         | 2      | 0      | -1     | 1      | 0      | 4    |
| \(s_3\)         | 0.5    | 0      | 0.5    | 0      | 1      | 7    |
| \(Z\)           | -0.5   | 0      | 1      | 0      | 0      | 8    |

### Step 4: Check for optimality

There are no negative coefficients in the objective function row (Z-row), so the current solution is optimal.

### Step

Using simplex method
Max z = x1+2x2
Subject to -x1+2x2<8, x1+2x2<12, x1-x2<3, x1, x2≥0

This detailed solution shows how to solve a linear programming problem using the Simplex Method. Maximize Z = x1 + 2x2 subject to several constraints. Includes key concepts like slack variables, pivot operations, and the Simplex tableau method. Suitable for college-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation