Maximize P = 4x + 20y subject to these constraints:
2x + 15y ≤ 700
5x + 10y ≤ 1150
5x + 5y ≤ 1000
4x + 15y ≤ 980
x ≥ 0, y ≥ 0
Maximum value for P = ?.
This value of P occurs when x = ? and y = ?
If u1, u2, u3, and u4 represent the slack variables in the four counstraints (u1 the slack variable in the top constraint listed above, u2 for the second, etc.), then give the values for each of the four slack variables when the optimal solution is reached:
u1 = ?
u2 = ?
u3 = ?
u4 = ?
Maximum value for P = 1150
This value of P occurs when x = 200 and y = 50
u1 = 0
u2 = 0
u3 = 150
u4 = 50
To solve this linear programming problem, we can use the simplex method. Here are the steps to find the maximum value for P and the corresponding values for x and y:
Step 1: Start with the initial tableau:
```
| 2x + 15y ≤ 700 |
| 5x + 10y ≤ 1150 |
| 5x + 5y ≤ 1000 |
| 4x + 15y ≤ 980 |
| x ≥ 0, y ≥ 0 |
---------------------------------
P | 4x + 20y - 0P = 0 |
```
Step 2: Convert the inequalities to equations by introducing slack variables. Let's introduce u1, u2, u3, and u4 as slack variables:
```
| 2x + 15y + u1 = 700 |
| 5x + 10y + u2 = 1150 |
| 5x + 5y + u3 = 1000 |
| 4x + 15y + u4 = 980 |
| x ≥ 0, y ≥ 0 |
----------------------------------
P | 4x + 20y - 0P = 0 |
```
Step 3: Apply the simplex method to find the maximum value for P and the corresponding values for x and y:
```
| 2 15 1 0 0 | 700 |
| 5 10 0 1 0 | 1150 |
| 5 5 0 0 1 | 1000 |
| 4 15 0 0 0 | 980 |
| -4 -20 0 0 0 | 0 |
---------------------------------
P | -4 -20 0 0 0 | 0 |
```
Step 4: Choose the most negative coefficient for P, which is -20. Pivot the tableau to make this coefficient positive:
```
| 2 7/3 1/3 0 0 | 460/3 |
| 5 10 0 1 0 | 1150 |
| 5 35/12 5/12 0 1 | 250/3 |
| 4 15 0 0 0 | 980 |
| 4 20/3 0 0 0 | 0 |
----------------------------------------
P | 0 0 0 0 0 | 100 |
```
Step 5: Repeat the process until there are no more negative coefficients for P:
```
| 2 7/3 -1/15 0 0 | 170 |
| 5 20/3 1/15 1 0 | 2100/13 |
| 5 25/12 2/15 0 1 | 190/3 |
| 4 25/3 -1/15 0 0 | 740/3 |
| 4 5/3 0 0 0 | 40/3 |
-------------------------------------------
P | 0 0 4/15 0 0 | 100 |
```
Step 6: The optimal solution is reached when all coefficients for P are non-negative. In this case, the maximum value for P is 100 and it occurs when x = 0 and y = 20.
So, the answers to your questions are:
- Maximum value for P = 100.
- The value of P occurs when x = 0 and y = 20.
- The values for the slack variables u1, u2, u3, and u4 when the optimal solution is reached are:
u1 = 170
u2 = 2100/13
u3 = 190/3
u4 = 740/3
To maximize the objective function P = 4x + 20y subject to the given constraints, we can solve this linear programming problem using the Simplex Method. Here's how to do it step by step:
Step 1: Write the constraint inequalities in standard form:
2x + 15y + u1 = 700
5x + 10y + u2 = 1150
5x + 5y + u3 = 1000
4x + 15y + u4 = 980
Step 2: Set up the initial simplex tableau:
| 2 15 1 0 0 0 | 700 |
| 5 10 0 1 0 0 | 1150 |
Tableau: | 5 5 0 0 1 0 | 1000 |
| 4 15 0 0 0 1 | 980 |
| -4 -20 0 0 0 0 | 0 |
Step 3: Choose the most negative coefficient (-20) in the bottom row for the entering variable. The entering variable is y.
Step 4: Calculate the ratios (divide the right-hand side by the corresponding coefficient of the entering variable) for each row where the corresponding coefficient of the entering variable is positive to determine the leaving variable. The smallest non-negative ratio corresponds to the leaving variable. In this case, it's row 2, so the leaving variable is u2.
Step 5: Perform row operations (Gaussian elimination) to make the pivot element (10) equal to 1 and clear the other elements in the column:
| 2 15 1 0 0 0 | 700 |
| 0 -2 -5 1 0 0 | 50 |
Tableau: | 5 5 0 0 1 0 | 1000 |
| 4 15 0 0 0 1 | 980 |
| 0 0 20 0 0 0 | 1000 |
Step 6: Calculate the pivot column (the most negative coefficient in the bottom row excluding the last column). The most negative coefficient is -5, which means x is the entering variable.
Step 7: Calculate the ratios for each row where the corresponding coefficient of the entering variable is positive to determine the leaving variable. The smallest non-negative ratio corresponds to the leaving variable. In this case, it's row 4, so the leaving variable is u4.
Step 8: Perform row operations:
| 2 15 1 0 0 0 | 700 |
| -8 -7 0 1 0 1 | -700 |
Tableau: | 5 5 0 0 1 0 | 1000 |
| 0 1 -20 0 0 0 | -1000 |
| 0 -15 20 0 0 0 | 9600 |
Step 9: The unboundedness test indicates that the problem is unbounded. Therefore, there is no finite maximum value for P.
However, if the problem was bounded, we would continue with additional iterations until we reach an optimal solution.
Regarding the values of x and y when the optimal solution is reached, we don't have a feasible solution in this case since the problem is unbounded.
As for the values of the slack variables u1, u2, u3, and u4 when the optimal solution is reached, we can't determine these values since the problem is unbounded.