Math Problem Statement

Using simplex method

Max z = x1+2x2

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

Solution

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=x1+2x2Z = x_1 + 2x_2

Subject to:

-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

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Simplex Method
Optimization

Formulas

Simplex tableau method
Pivot operation in simplex
Slack variables transformation

Theorems

Simplex algorithm for optimization

Suitable Grade Level

College/University level