Answer :
To set up the initial simplex tableau for this problem, we need to consider the constraints given by the availability of nutrients and the growth values from each type of food. Let's break it down step by step:
### Step 1: Identify the Variables
The variables represent the number of kilograms of each food type to be produced:
- [tex]\( x_1 \)[/tex]: Kilograms of food type [tex]\( P \)[/tex]
- [tex]\( x_2 \)[/tex]: Kilograms of food type [tex]\( Q \)[/tex]
- [tex]\( x_3 \)[/tex]: Kilograms of food type [tex]\( R \)[/tex]
- [tex]\( x_4 \)[/tex]: Kilograms of food type [tex]\( S \)[/tex]
### Step 2: Write the Objective Function
The objective is to maximize the total growth value, which is given by:
[tex]\[ z = 100x_1 + 80x_2 + 70x_3 + 50x_4 \][/tex]
### Step 3: Set Up Constraints
Based on the nutrient availability and food composition, the constraints are:
- For nutrient [tex]\( A \)[/tex]: [tex]\( 0.375x_3 + 0.125x_4 \leq 600 \)[/tex]
- For nutrient [tex]\( B \)[/tex]: [tex]\( 0.5x_2 + 0.25x_3 + 0.875x_4 \leq 700 \)[/tex]
- For nutrient [tex]\( C \)[/tex]: [tex]\( x_1 + 0.5x_2 + 0.375x_3 \leq 300 \)[/tex]
### Step 4: Convert to Standard Form with Slack Variables
To convert each inequality into an equation, we introduce slack variables:
- [tex]\( s_1 \)[/tex] for nutrient [tex]\( A \)[/tex]: [tex]\( 0.375x_3 + 0.125x_4 + s_1 = 600 \)[/tex]
- [tex]\( s_2 \)[/tex] for nutrient [tex]\( B \)[/tex]: [tex]\( 0.5x_2 + 0.25x_3 + 0.875x_4 + s_2 = 700 \)[/tex]
- [tex]\( s_3 \)[/tex] for nutrient [tex]\( C \)[/tex]: [tex]\( x_1 + 0.5x_2 + 0.375x_3 + s_3 = 300 \)[/tex]
### Step 5: Construct the Initial Simplex Tableau
The tableau is a way to organize the coefficients of the variables in these equations and the objective function. It looks like this:
[tex]\[
\begin{array}{cccccccc}
x_1 & x_2 & x_3 & x_4 & s_1 & s_2 & s_3 & \text{Solution} \\
\hline
0 & 0 & 0.375 & 0.125 & 1 & 0 & 0 & 600 \\
0 & 0.5 & 0.25 & 0.875 & 0 & 1 & 0 & 700 \\
1 & 0.5 & 0.375 & 0 & 0 & 0 & 1 & 300 \\
\hline
100 & 80 & -70 & -50 & 0 & 0 & 0 & 0
\end{array}
\][/tex]
In this tableau:
- The first three rows represent the constraints with slack variables added to convert inequalities to equalities.
- The last row represents the negative of the coefficients of the objective function terms, as it is a maximization problem. The constant term is zero because we haven't produced any food yet and therefore have no growth value.
This tableau is ready for the process of the simplex method to find the optimal solution.
### Step 1: Identify the Variables
The variables represent the number of kilograms of each food type to be produced:
- [tex]\( x_1 \)[/tex]: Kilograms of food type [tex]\( P \)[/tex]
- [tex]\( x_2 \)[/tex]: Kilograms of food type [tex]\( Q \)[/tex]
- [tex]\( x_3 \)[/tex]: Kilograms of food type [tex]\( R \)[/tex]
- [tex]\( x_4 \)[/tex]: Kilograms of food type [tex]\( S \)[/tex]
### Step 2: Write the Objective Function
The objective is to maximize the total growth value, which is given by:
[tex]\[ z = 100x_1 + 80x_2 + 70x_3 + 50x_4 \][/tex]
### Step 3: Set Up Constraints
Based on the nutrient availability and food composition, the constraints are:
- For nutrient [tex]\( A \)[/tex]: [tex]\( 0.375x_3 + 0.125x_4 \leq 600 \)[/tex]
- For nutrient [tex]\( B \)[/tex]: [tex]\( 0.5x_2 + 0.25x_3 + 0.875x_4 \leq 700 \)[/tex]
- For nutrient [tex]\( C \)[/tex]: [tex]\( x_1 + 0.5x_2 + 0.375x_3 \leq 300 \)[/tex]
### Step 4: Convert to Standard Form with Slack Variables
To convert each inequality into an equation, we introduce slack variables:
- [tex]\( s_1 \)[/tex] for nutrient [tex]\( A \)[/tex]: [tex]\( 0.375x_3 + 0.125x_4 + s_1 = 600 \)[/tex]
- [tex]\( s_2 \)[/tex] for nutrient [tex]\( B \)[/tex]: [tex]\( 0.5x_2 + 0.25x_3 + 0.875x_4 + s_2 = 700 \)[/tex]
- [tex]\( s_3 \)[/tex] for nutrient [tex]\( C \)[/tex]: [tex]\( x_1 + 0.5x_2 + 0.375x_3 + s_3 = 300 \)[/tex]
### Step 5: Construct the Initial Simplex Tableau
The tableau is a way to organize the coefficients of the variables in these equations and the objective function. It looks like this:
[tex]\[
\begin{array}{cccccccc}
x_1 & x_2 & x_3 & x_4 & s_1 & s_2 & s_3 & \text{Solution} \\
\hline
0 & 0 & 0.375 & 0.125 & 1 & 0 & 0 & 600 \\
0 & 0.5 & 0.25 & 0.875 & 0 & 1 & 0 & 700 \\
1 & 0.5 & 0.375 & 0 & 0 & 0 & 1 & 300 \\
\hline
100 & 80 & -70 & -50 & 0 & 0 & 0 & 0
\end{array}
\][/tex]
In this tableau:
- The first three rows represent the constraints with slack variables added to convert inequalities to equalities.
- The last row represents the negative of the coefficients of the objective function terms, as it is a maximization problem. The constant term is zero because we haven't produced any food yet and therefore have no growth value.
This tableau is ready for the process of the simplex method to find the optimal solution.
