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Minimize using the simplex method: Z = 8x1 + 10x2 + 7x3 Subject to: x1 + 3x2 + 2x3 ≤ 10 x1  5x2  x3 ≥ 8 x1, x2, x3 ≥ 0
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more than 3 weeks ago
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#mathematics
#optimization
Minimize using the simplex method: Z = 8x1 + 10x2 + 7x3 Subject to: x1 + 3x2 + 2x3 ≤ 10 x1  5x2  x3 ≥ 8 x1, x2, x3 ≥ 0
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ChatGPT
more than 3 weeks ago
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To minimize Z = 8x1 + 10x2 + 7x3, subject to the constraints:
1. x1 + 3x2 + 2x3 ≤ 10
2. x1  5x2  x3 ≥ 8
3. x1, x2, x3 ≥ 0
We first convert the inequalities to equations by introducing slack variables:
1. x1 + 3x2 + 2x3 + x4 = 10
2. x1  5x2  x3  x5 = 8
The initial simplex tableau is:
```
 x1  x2  x3  x4  x5  RHS 

 1  3  2  1  0  10 
 1  5  1  0  1  8 
 8  10  7  0  0  0 
```
We choose x1 as the entering variable and x5 as the leaving variable. Perform row operations to pivot:
```
 x1  x2  x3  x4  x5  RHS 

 0  2  1  1  1  2 
 1  3  2  1  0  10 
 0  30  23  8  8  80 
```
The next pivot is x2 entering and x4 leaving:
```
 x1  x2  x3  x4  x5  RHS 

 0  0  5  2  3  8 
 1  0  4  5  3  2 
 0  1  23/30  8/30  8/30  8/3 
```
The optimal solution is x1 = 2, x2 = 8/3, x3 = 8/3 with Z = 8(2) + 10(8/3) + 7(8/3) = 64/3.
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Google Gemini AI
more than 3 weeks ago
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Step 1: Convert to Standard Form
Introduce slack variables s1 and s2 to convert the inequality constraints to equalities:
x1 3x2 2x3 + s1 = 10
x1  5x2  x3 + s2 = 8
x1, x2, x3, s1, s2 ≥ 0
Step 2: Construct the Initial Simplex Tableau

Variable

x1

x2

x3

s1

s2

RHS



Z
 8  10  7  0  0  0 

s1
 1  3  2  1  0  10 

s2
 1  5  1  0  1  8 
Step 3: Choose the Entering Variable
The variable with the most negative coefficient in the Zrow is x1, so x1 enters the basis.
Step 4: Choose the Leaving Variable
To determine the leaving variable, calculate the ratios of the RHS to the coefficients of x1 in the constraint rows:
Ratio for s1: 10 / 1 = 10
Ratio for s2: 8 / 1 = 8
The smallest ratio is 8, so s2 leaves the basis.
Step 5: Perform the Pivot Operation
Divide the row of s2 by its pivot element (1) to get:

Variable

x1

x2

x3

s1

s2

RHS



Z
 8  10  7  0  0  0 

s1
 1  3  2  1  0  10 

x1
 1  5  1  0  1  8 
Step 6: Repeat Steps 35 until an Optimal Solution is Reached
x2 enters the basis, s1 leaves the basis.
x3 enters the basis, x2 leaves the basis.
The final simplex tableau is:

Variable

x1

x2

x3

s1

s2

RHS



Z
 0  0  0  0  0  68 

s1
 0  0  1  1  0  2 

x2
 0  0  2  1  0  4 

x3
 0  0  2  0  1  10 
Optimal Solution:
Z = 68
x1 = 0
x2 = 4
x3 = 10
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