Using the simplex method, solve the following linear programming problem:
Maximize: P= 5x+2y
Subject to: 4x+3y < or equal to 30
2x-3y < or equal to 6
x > or equal to 0, y > or equal to 0.
To solve this linear programming problem using the simplex method, follow these steps:
Step 1: Convert the problem to standard form:
- Rewrite the objective function as a maximization problem by changing the sign of the coefficients.
- Add slack variables (s1 and s2) for each inequality constraint to convert them into equality constraints.
- Add non-negativity constraints for the slack variables.
The problem in standard form:
Maximize: P = -5x - 2y + 0s₁ + 0s₂
Subject to: 4x + 3y + s₁ = 30
2x - 3y + s₂ = 6
x ≥ 0, y ≥ 0, s₁ ≥ 0, s₂ ≥ 0
Step 2: Create the initial simplex tableau:
- Convert the standard form equations into a tableau.
- Set up the initial simplex tableau by arranging the variables and coefficients in a matrix form.
The initial tableau:
| Basis | x | y | s₁ | s₂ | RHS |
|-------|---|---|----|----|-----|
| s₁ | 4 | 3 | 1 | 0 | 30 |
| s₂ | 2 | -3| 0 | 1 | 6 |
| Z | -5| -2| 0 | 0 | 0 |
Step 3: Select the pivot column and pivot row:
- Choose the most negative coefficient in the bottom row as the pivot column.
- Divide the right-hand side (RHS) of each equation by the corresponding pivot column coefficient to determine the pivot row.
- Determine the pivot element at the intersection of the pivot column and pivot row.
In this case, the most negative coefficient is -5. The pivot row is determined by dividing the RHS of each equation by the coefficients under the pivot column:
Pivot row for s₁: 30/4 = 7.5
Pivot row for s₂: 6/2 = 3
Since 3 is smaller, the pivot row is s₂.
Step 4: Perform row operations to make the pivot element equal to 1 and other elements in the pivot column equal to 0:
- Divide the pivot row by the pivot element to make it equal to 1.
- Perform row operations to make other elements in the pivot column equal to 0.
The updated tableau:
| Basis | x | y | s₁ | s₂ | RHS |
|-------|---|---|----|----|-----|
| s₁ | 7 | 0 | 1 | 1 | 27 |
| x | 1 | -1| 0 | 2/3| 2 |
| Z | 5 | 0 | 0 | -10| 30 |
Step 5: Repeat steps 3 and 4 until the bottom row coefficients are all non-negative:
- Select the most negative coefficient in the bottom row as the new pivot column.
- Determine the pivot row using the same process as before.
- Update the tableau with row operations.
The updated tableau:
| Basis | x | y | s₁ | s₂ | RHS |
|-------|---|---|----|----|-----|
| y | 7/2| 0 | 1/2 | 1/2 | 27/2 |
| x | 0 | -1| -1/7 | 8/7| 23/7 |
| Z | 0 | 0 | 10/7 | -65/7| 95/7 |
Step 6: Repeat steps 3 and 4 until the bottom row coefficients are all non-negative.
Since all the coefficients in the bottom row are non-negative, the optimal solution is found.
Step 7: Interpret the tableau:
- The Z row represents the objective function, and the values under the RHS column represent the optimal values for each variable.
- In this case, P (the objective function) has a maximum value of 95/7.
- The optimal values for x, y, s₁, and s₂ are 23/7, 0, 27/2, and 0, respectively.
Therefore, the maximum value of the objective function P is 95/7, and it occurs when x = 23/7, y = 0.