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To solve the given linear programming problem using the simplex method, we need to convert it into its standard form, which involves the following steps:
1. Formulate the Objective Function and Constraints:
[tex]\[ \begin{array}{ll} \text{Maximize} & z = 6x_1 + 3x_2 + x_3 \\ \text{subject to} & 5x_1 + 5x_2 + x_3 \leq 25 \\ & x_1 + 3x_2 + 3x_3 \leq 18 \\ & x_1 \geq 0, x_2 \geq 0, x_3 \geq 0 \end{array} \][/tex]
2. Introduce Slack Variables:
To convert the inequalities into equalities, we introduce slack variables [tex]\( s_1 \)[/tex] and [tex]\( s_2 \)[/tex]:
[tex]\[ \begin{array}{ll} 5x_1 + 5x_2 + x_3 + s_1 = 25 \\ x_1 + 3x_2 + 3x_3 + s_2 = 18 \\ s_1 \geq 0, s_2 \geq 0 \end{array} \][/tex]
3. Set Up the Initial Simplex Tableau:
The initial simplex tableau is:
[tex]\[ \begin{array}{c|cccccc|c} \text{Basic} & x_1 & x_2 & x_3 & s_1 & s_2 & \text{RHS} & \\ \hline s_1 & 5 & 5 & 1 & 1 & 0 & 25 \\ s_2 & 1 & 3 & 3 & 0 & 1 & 18 \\ \hline \text{Z} & -6 & -3 & -1 & 0 & 0 & 0 \\ \end{array} \][/tex]
4. Perform the Simplex Algorithm:
- Identify the entering variable (most negative coefficient in the Z row):
- Here, [tex]\( -6 \)[/tex] is the most negative, so [tex]\( x_1 \)[/tex] enters the basis.
- Determine the leaving variable using the minimum ratio test:
- For [tex]\( s_1 \)[/tex]: [tex]\( \frac{25}{5} = 5 \)[/tex]
- For [tex]\( s_2 \)[/tex]: [tex]\( \frac{18}{1} = 18 \)[/tex]
- Since [tex]\( 5 \)[/tex] is smaller, [tex]\( s_1 \)[/tex] leaves the basis.
- Perform the pivot operation to update the tableau. This involves making [tex]\( x_1 \)[/tex]’s entry in the pivot row equal to 1 and other entries in its column to 0.
After performing all necessary pivot operations and iterations (not shown in full detail here for brevity), we reach the optimal tableau:
5. Optimal Solution:
From the final tableau, we find:
[tex]\[ \begin{array}{c|cccccc|c} \text{Basic} & x_1 & x_2 & x_3 & s_1 & s_2 & \text{RHS} \\ \hline x_1 & 1 & 0 & 0 & 0.2 & -1 & 5 \\ s_2 & 0 & 1 & 0 & -0.2 & 0.2 & 3 \\ \hline \text{Z} & 0 & 0 & 0 & 1.2 & -6 & 30 \\ \end{array} \][/tex]
Therefore, the optimal solution is:
[tex]\[ \begin{cases} x_1 = 5 \\ x_2 = 0 \\ x_3 = 0 \\ s_1 = 0 \\ s_2 = 3 \\ \end{cases} \][/tex]
The maximum value of [tex]\( z \)[/tex] is [tex]\( 30 \)[/tex].
So the final choice is:
A. The maximum is [tex]\( 30 \)[/tex] when [tex]\( x_1 = 5 \)[/tex], [tex]\( x_2 = 0 \)[/tex], [tex]\( x_3 = 0 \)[/tex], [tex]\( s_1 = 0 \)[/tex], and [tex]\( s_2 = 3 \)[/tex].
1. Formulate the Objective Function and Constraints:
[tex]\[ \begin{array}{ll} \text{Maximize} & z = 6x_1 + 3x_2 + x_3 \\ \text{subject to} & 5x_1 + 5x_2 + x_3 \leq 25 \\ & x_1 + 3x_2 + 3x_3 \leq 18 \\ & x_1 \geq 0, x_2 \geq 0, x_3 \geq 0 \end{array} \][/tex]
2. Introduce Slack Variables:
To convert the inequalities into equalities, we introduce slack variables [tex]\( s_1 \)[/tex] and [tex]\( s_2 \)[/tex]:
[tex]\[ \begin{array}{ll} 5x_1 + 5x_2 + x_3 + s_1 = 25 \\ x_1 + 3x_2 + 3x_3 + s_2 = 18 \\ s_1 \geq 0, s_2 \geq 0 \end{array} \][/tex]
3. Set Up the Initial Simplex Tableau:
The initial simplex tableau is:
[tex]\[ \begin{array}{c|cccccc|c} \text{Basic} & x_1 & x_2 & x_3 & s_1 & s_2 & \text{RHS} & \\ \hline s_1 & 5 & 5 & 1 & 1 & 0 & 25 \\ s_2 & 1 & 3 & 3 & 0 & 1 & 18 \\ \hline \text{Z} & -6 & -3 & -1 & 0 & 0 & 0 \\ \end{array} \][/tex]
4. Perform the Simplex Algorithm:
- Identify the entering variable (most negative coefficient in the Z row):
- Here, [tex]\( -6 \)[/tex] is the most negative, so [tex]\( x_1 \)[/tex] enters the basis.
- Determine the leaving variable using the minimum ratio test:
- For [tex]\( s_1 \)[/tex]: [tex]\( \frac{25}{5} = 5 \)[/tex]
- For [tex]\( s_2 \)[/tex]: [tex]\( \frac{18}{1} = 18 \)[/tex]
- Since [tex]\( 5 \)[/tex] is smaller, [tex]\( s_1 \)[/tex] leaves the basis.
- Perform the pivot operation to update the tableau. This involves making [tex]\( x_1 \)[/tex]’s entry in the pivot row equal to 1 and other entries in its column to 0.
After performing all necessary pivot operations and iterations (not shown in full detail here for brevity), we reach the optimal tableau:
5. Optimal Solution:
From the final tableau, we find:
[tex]\[ \begin{array}{c|cccccc|c} \text{Basic} & x_1 & x_2 & x_3 & s_1 & s_2 & \text{RHS} \\ \hline x_1 & 1 & 0 & 0 & 0.2 & -1 & 5 \\ s_2 & 0 & 1 & 0 & -0.2 & 0.2 & 3 \\ \hline \text{Z} & 0 & 0 & 0 & 1.2 & -6 & 30 \\ \end{array} \][/tex]
Therefore, the optimal solution is:
[tex]\[ \begin{cases} x_1 = 5 \\ x_2 = 0 \\ x_3 = 0 \\ s_1 = 0 \\ s_2 = 3 \\ \end{cases} \][/tex]
The maximum value of [tex]\( z \)[/tex] is [tex]\( 30 \)[/tex].
So the final choice is:
A. The maximum is [tex]\( 30 \)[/tex] when [tex]\( x_1 = 5 \)[/tex], [tex]\( x_2 = 0 \)[/tex], [tex]\( x_3 = 0 \)[/tex], [tex]\( s_1 = 0 \)[/tex], and [tex]\( s_2 = 3 \)[/tex].
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