## Symbolic Representation

To solve the given optimization problem, we first need to convert the natural language description into a symbolic representation.

Let's define the variables:
- $x_1$ = Number of apple pies
- $x_2$ = Number of blueberry pies

The objective is to maximize profit. Given that the profit per apple pie is $5 and the profit per blueberry pie is $6, the objective function can be written as:

Maximize: $5x_1 + 6x_2$

The constraints based on the given information are:
1. Time constraint: It takes 30 minutes to make one apple pie and 40 minutes to make a blueberry pie, with a maximum of 4500 minutes per week.
   - $30x_1 + 40x_2 \leq 4500$

2. Apple pies production limit: They can make at most 70 apple pies.
   - $x_1 \leq 70$

3. Blueberry pies production limit: They can make at most 60 blueberry pies.
   - $x_2 \leq 60$

4. Non-negativity constraint: The number of pies cannot be negative.
   - $x_1 \geq 0, x_2 \geq 0$

## Symbolic Representation in JSON Format

```json
{
    'sym_variables': [('x1', 'apple pies'), ('x2', 'blueberry pies')],
    'objective_function': '5*x1 + 6*x2',
    'constraints': [
        '30*x1 + 40*x2 <= 4500',
        'x1 <= 70',
        'x2 <= 60',
        'x1 >= 0',
        'x2 >= 0'
    ]
}
```

## Gurobi Code in Python

```python
import gurobi

def solve_pie_production():
    # Create a new model
    model = gurobi.Model()

    # Define variables
    x1 = model.addVar(lb=0, name="apple_pies")  # Number of apple pies
    x2 = model.addVar(lb=0, name="blueberry_pies")  # Number of blueberry pies

    # Objective function: Maximize profit
    model.setObjective(5 * x1 + 6 * x2, gurobi.GRB.MAXIMIZE)

    # Constraints
    model.addConstr(30 * x1 + 40 * x2 <= 4500, name="time_constraint")
    model.addConstr(x1 <= 70, name="apple_pies_limit")
    model.addConstr(x2 <= 60, name="blueberry_pies_limit")

    # Optimize the model
    model.optimize()

    # Print the results
    if model.status == gurobi.GRB.OPTIMAL:
        print(f"Optimal solution: Apple pies = {x1.varValue}, Blueberry pies = {x2.varValue}")
        print(f"Maximum profit: ${5 * x1.varValue + 6 * x2.varValue}")
    else:
        print("No optimal solution found.")

solve_pie_production()
```