To solve the given optimization problem, we first need to translate the natural language description into a symbolic representation. This involves defining variables and formulating the objective function and constraints algebraically.

Let's define:
- $x_1$ as the number of blueberry smoothies made,
- $x_2$ as the number of chocolate smoothies made.

The profit per blueberry smoothie is $2, and the profit per chocolate smoothie is $3. Therefore, the objective function to maximize profit can be written as:
\[ 2x_1 + 3x_2 \]

The constraints are based on the time available for preparation and blending:
- Preparation time: $3x_1 + 5x_2 \leq 1000$ (since each blueberry smoothie requires 3 minutes of preparation, each chocolate smoothie requires 5 minutes, and there are 1000 minutes available),
- Blending time: $2x_1 + x_2 \leq 750$ (since each blueberry smoothie requires 2 minutes of blending, each chocolate smoothie requires 1 minute, and there are 750 minutes available).

Additionally, we have non-negativity constraints since the number of smoothies cannot be negative:
- $x_1 \geq 0$,
- $x_2 \geq 0$.

The symbolic representation of the problem can be summarized as follows:
```json
{
    'sym_variables': [('x1', 'number of blueberry smoothies'), ('x2', 'number of chocolate smoothies')],
    'objective_function': '2*x1 + 3*x2',
    'constraints': ['3*x1 + 5*x2 <= 1000', '2*x1 + x2 <= 750', 'x1 >= 0', 'x2 >= 0']
}
```

To solve this problem using Gurobi in Python, we'll use the following code:

```python
from gurobipy import *

# Create a new model
m = Model("Smoothies_Optimization")

# Define variables
x1 = m.addVar(name='blueberry_smoothies', vtype=GRB.CONTINUOUS, lb=0)
x2 = m.addVar(name='chocolate_smoothies', vtype=GRB.CONTINUOUS, lb=0)

# Set the objective function
m.setObjective(2*x1 + 3*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(3*x1 + 5*x2 <= 1000, name='preparation_time')
m.addConstr(2*x1 + x2 <= 750, name='blending_time')

# Optimize the model
m.optimize()

# Print results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Number of blueberry smoothies: {x1.x}")
    print(f"Number of chocolate smoothies: {x2.x}")
    print(f"Maximum profit: {m.objVal}")
else:
    print("No optimal solution found")
```