Maximize P=60x+50y.
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⎪x+y≤80
⎪
⎪5x+10y≤560
⎪
⎪50x+20y≤1600
⎪
⎪x≥0
⎨
⎪y≥0
⎪
⎪
⎪
⎪
⎪
⎪
⎩
Use the simplex method to solve the problem. Use s, t, and u as your slack variables for the first, second, and third inequalities respectively. Use the final simplex tableau to identify the values below.
x =
y =
s =
t =
u =
P =
http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=1e692c6f72587b2cbd3e7be018fd8960&title=Linear%20Programming%20Calculator&theme=blue
max = 3220 at (12,50)
For instructions using Simplex with pivots etc see
http://en.wikipedia.org/wiki/Simplex_algorithm
To solve the linear programming problem using the simplex method, we start by setting up the initial simplex tableau.
First, we rewrite the inequalities in equation form:
x + y + s = 80
5x + 10y + t = 560
50x + 20y + u = 1600
We convert the objective function to an equation by subtracting P from both sides:
60x + 50y - P = 0
Now we set up the initial simplex tableau:
---------------------
| x | y | s | t | u | RHS |
---------------------
| 1 | 1 | 1 | 0 | 0 | 80 |
| 5 | 10| 0 | 1 | 0 | 560|
| 50| 20| 0 | 0 | 1 | 1600|
| 60| 50| 0 | 0 | 0 | 0 |
---------------------
To start the simplex iteration, we select the most negative coefficient in the bottom row, which is -50 in this case. We choose the column with the smallest ratio from the positive entries in the respective column. Since the ratio for column y is the smallest (80/1 = 80), we proceed with column y.
Next, we perform row operations to make all other entries in column y zero, except for the pivot element (1 in row 1). The pivot operation is as follows:
R1 = R1 - R2/10
R3 = R3 - 2R2
R4 = R4 - 5R2
The updated tableau is:
---------------------
| x | y | s | t | u | RHS |
---------------------
| 0 | 1 | 1/10 | -1/10 | 0 | 8 |
| 5 | 0 |-1/10 | 11/10 | 0 |512 |
| 40| 0 |-1 | 2 | 1 |480 |
| -10| 0 | 1/2 | -5/2 | 0 | -320 |
---------------------
Now, we select the column with the most negative entry in the bottom row, which is -5/2. The smallest ratio from the positive entries in the respective column is 512/11, so we proceed with column t.
We perform row operations to make all other entries in column t zero, except for the pivot element (1/2 in row 4). The pivot operation is as follows:
R1 = R1 - (11/10)R4/5
R2 = R2 - (11/10)R4
R3 = R3 + (5/2)R4
R4 = 2R4/5
The updated tableau is:
---------------------
| x | y | s | t | u | RHS |
---------------------
| 0 | 1 | 0 | 1 | 0 | 12 |
| 5 | 0 | -3/10 | 0 | 0 |476 |
| 40| 0 | -7/5 | 0 | 1 |536 |
| 0| 0 | 1 | 0 | 0 | 128 |
---------------------
Now, there are no negative entries in the bottom row, so the optimal solution has been reached.
From the final simplex tableau:
x = 12
y = 0
s = 0
t = 0
u = 128
P = 60(12) + 50(0) = 720
To solve this problem using the simplex method, we can follow these steps:
1. Convert the problem into standard form by introducing slack variables.
The given inequalities can be rewritten as:
x + y + s = 80 (First constraint)
5x + 10y + t = 560 (Second constraint)
50x + 20y + u = 1600 (Third constraint)
2. Set up the initial simplex tableau:
| Coefficients | x | y | s | t | u | RHS |
|----------------|-----|-----|-----|-----|-----|------|
| P-row | -60 | -50 | 0 | 0 | 0 | 0 |
| s-row | 1 | 1 | 1 | 0 | 0 | 80 |
| t-row | 5 | 10 | 0 | 1 | 0 | 560 |
| u-row | 50 | 20 | 0 | 0 | 1 | 1600 |
3. Apply the simplex method to optimize P:
- Select the most negative coefficient of the P-row as the pivot column (in this case, -60).
- Choose the pivot row by finding the ratio of the RHS value to the corresponding pivot column value. Select the smallest ratio row (in this case, the s-row).
- Divide the pivot row by the pivot element (1 in this case) to make the pivot element equal to 1.
- Update the other rows by performing row operations to make all the other pivot column elements equal to 0.
- Repeat these steps until all the coefficients in the P-row are non-negative.
4. Repeat the process until all the coefficients in the P-row are non-negative. Once all the coefficients are non-negative, the solution is optimal.
Once you have completed the iterations and identified the final simplex tableau, you can read the values below:
x = value in the x-column of the optimal tableau
y = value in the y-column of the optimal tableau
s = value in the s-column of the optimal tableau
t = value in the t-column of the optimal tableau
u = value in the u-column of the optimal tableau
P = value in the RHS column in the P-row of the optimal tableau. This will give you the maximum value of P.
Please note that without the specific data for the RHS values, it is not possible to determine the exact values of x, y, s, t, and u, as well as the value of P in this problem.