To solve the given optimization problem, we first need to convert the natural language description into a symbolic representation. This involves defining variables for the quantities of large and small sandwiches, formulating the objective function that represents the total profit, and expressing the constraints related to preparation and toasting times.

Let's denote:
- \(x_1\) as the number of large sandwiches,
- \(x_2\) as the number of small sandwiches.

The objective function, which is the total profit, can be represented as:
\[5x_1 + 3.50x_2\]

Given that each large sandwich takes 4 minutes of preparation and each small sandwich takes 3 minutes, and there are 1000 minutes available for preparation, we have the constraint:
\[4x_1 + 3x_2 \leq 1000\]

For toasting, with 5 minutes per large sandwich and 4 minutes per small sandwich, and 1200 minutes available, the constraint is:
\[5x_1 + 4x_2 \leq 1200\]

Additionally, \(x_1\) and \(x_2\) must be non-negative since they represent quantities of sandwiches.

The symbolic representation in JSON format is:

```json
{
    'sym_variables': [('x1', 'number of large sandwiches'), ('x2', 'number of small sandwiches')],
    'objective_function': '5*x1 + 3.50*x2',
    'constraints': ['4*x1 + 3*x2 <= 1000', '5*x1 + 4*x2 <= 1200', '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("Sandwich_Problem")

# Define variables
x1 = m.addVar(name="large_sandwiches", vtype=GRB.CONTINUOUS, lb=0)
x2 = m.addVar(name="small_sandwiches", vtype=GRB.CONTINUOUS, lb=0)

# Set the objective function
m.setObjective(5*x1 + 3.50*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(4*x1 + 3*x2 <= 1000, name="preparation_time")
m.addConstr(5*x1 + 4*x2 <= 1200, name="toasting_time")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Number of large sandwiches: {x1.x}")
    print(f"Number of small sandwiches: {x2.x}")
    print(f"Total profit: {m.objVal}")
else:
    print("No optimal solution found")
```