SOLUTION: Solve the following problem by simplex method. Max Z = 8x1 + 16x2 Subject to: x1 + x2  200 x2  125 3x1 + 6x2

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Question 1189862: Solve the following problem by simplex method.
Max Z = 8x1 + 16x2
Subject to: x1 + x2  200
x2  125
3x1 + 6x2  900
Where x1, x2  0 non-negativity condition



Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
Here's how to solve the given linear programming problem using the simplex method:
**1. Convert Inequalities to Equations:**
Introduce slack variables (s1, s2, s3) to convert the inequalities into equations:
* x1 + x2 + s1 = 200
* x2 + s2 = 125
* 3x1 + 6x2 + s3 = 900
**2. Set up the Initial Simplex Tableau:**
| | x1 | x2 | s1 | s2 | s3 | RHS |
|---|---|---|---|---|---|---|
| Z | -8 | -16 | 0 | 0 | 0 | 0 |
| | 1 | 1 | 1 | 0 | 0 | 200 |
| | 0 | 1 | 0 | 1 | 0 | 125 |
| | 3 | 6 | 0 | 0 | 1 | 900 |
**3. Identify the Pivot Column:**
The pivot column is the column with the most negative entry in the Z row. In this case, it's the x2 column (-16).
**4. Identify the Pivot Row:**
Divide the RHS values by the corresponding entries in the pivot column. The smallest non-negative result indicates the pivot row.
* 200 / 1 = 200
* 125 / 1 = 125
* 900 / 6 = 150
The pivot row is the second row.
**5. Pivot:**
The pivot element is the intersection of the pivot column and pivot row (1). Make the pivot element 1 (it already is) and all other elements in the pivot column 0.
| | x1 | x2 | s1 | s2 | s3 | RHS |
|---|---|---|---|---|---|---|
| Z | -8 | 0 | 0 | 16 | 0 | 2000 |
| | 1 | 0 | 1 | -1 | 0 | 75 |
| | 0 | 1 | 0 | 1 | 0 | 125 |
| | 3 | 0 | 0 | -6 | 1 | 150 |
**6. Repeat Steps 3-5:**
* Pivot Column: x1 (-8)
* Pivot Row: First Row (75/1 is the smallest)
* Pivot:
| | x1 | x2 | s1 | s2 | s3 | RHS |
|---|---|---|---|---|---|---|
| Z | 0 | 0 | 8 | 8 | 0 | 2600 |
| | 1 | 0 | 1 | -1 | 0 | 75 |
| | 0 | 1 | 0 | 1 | 0 | 125 |
| | 0 | 0 | -3 | -3 | 1 | -75 |
**7. Optimal Solution:**
Since there are no more negative entries in the Z row, the optimal solution is reached.
* x1 = 50
* x2 = 125
* Z = 2400
Therefore, the maximum value of Z is 2400 when x1 = 50 and x2 = 125.
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
Solve the following problem by simplex method.

Max Z = 8x1 + 16x2
Subject to: x1 + x2 <= 200
x2 <= 125
3x1 + 6x2 <= 900
Where x1, x2 > 0 (non-negativity condition)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        It can be solved mentally, using common sense only
        and without using the simplex method. 


Indeed, from the objective function form  Z = 8x1 + 16x2 
it is clear that it is profitable to keep x2 as large as possible due to constrains
and after determining x2 in this way to keep x1 as large as possible due to the constrains.


Doing this way, we take  x2 = 125  (maximum allowed by the constrain x2 <= 125).


Then we determine x1 =  = 50 from constrain 3x1 + 6x2 <= 900.



Then   = 8x1 + 16x2 = 8*50 + 16*125 = 2400.

Solved mentally.



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