## Problem Description and Symbolic Representation

The problem involves maximizing profit by determining the optimal number of blueberry and chocolate smoothies to produce given certain constraints on preparation and blending times.

### Symbolic Variables:
- $x_1$ = Number of blueberry smoothies
- $x_2$ = Number of chocolate smoothies

### Objective Function:
The profit per blueberry smoothie is $2, and the profit per chocolate smoothie is $3. Therefore, the objective function to maximize profit ($P$) is:
\[ P = 2x_1 + 3x_2 \]

### Constraints:
1. Preparation time: A blueberry smoothie requires 3 minutes, and a chocolate smoothie requires 5 minutes. The store has 1000 minutes available for preparation.
\[ 3x_1 + 5x_2 \leq 1000 \]

2. Blending time: A blueberry smoothie requires 2 minutes, and a chocolate smoothie requires 1 minute. The store has 750 minutes available for blending.
\[ 2x_1 + x_2 \leq 750 \]

3. Non-negativity constraint: The number of smoothies cannot be negative.
\[ x_1 \geq 0, x_2 \geq 0 \]

## Symbolic Representation in JSON Format

```json
{
    'sym_variables': [('x1', 'blueberry smoothies'), ('x2', 'chocolate smoothies')],
    'objective_function': '2*x1 + 3*x2',
    'constraints': [
        '3*x1 + 5*x2 <= 1000',
        '2*x1 + x2 <= 750',
        'x1 >= 0',
        'x2 >= 0'
    ]
}
```

## Gurobi Code in Python

```python
import gurobi

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

    # Define variables
    x1 = model.addVar(name="blueberry_smoothies", lb=0, vtype=gurobi.GRB.CONTINUOUS)
    x2 = model.addVar(name="chocolate_smoothies", lb=0, vtype=gurobi.GRB.CONTINUOUS)

    # Objective function: Maximize 2*x1 + 3*x2
    model.setObjective(2*x1 + 3*x2, gurobi.GRB.MAXIMIZE)

    # Constraints
    model.addConstr(3*x1 + 5*x2 <= 1000, name="preparation_time")
    model.addConstr(2*x1 + x2 <= 750, name="blending_time")

    # Solve the model
    model.optimize()

    # Print the solution
    if model.status == gurobi.GRB.OPTIMAL:
        print(f"Optimal solution: x1 = {x1.varValue}, x2 = {x2.varValue}")
        print(f"Maximum profit: {model.objVal}")
    else:
        print("The model is infeasible")

solve_smoothie_problem()
```