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

- $x_1$ as the number of batches of sausages produced,
- $x_2$ as the number of batches of burger patties produced.

The objective function aims to maximize profit. Given that the profit per batch of sausages is $200 and the profit per batch of burger patties is $250, the objective function can be represented algebraically as:

\[ \text{Maximize:} \quad 200x_1 + 250x_2 \]

The constraints are based on the time requirements for each machine. Since producing one batch of sausages requires 2 hours on the meat-grinder and one batch of burger patties requires 4 hours, and given that each machine can run for at most 3000 hours per year, we have:

- For the meat-grinder: $2x_1 + 4x_2 \leq 3000$
- For the meat-packer: $3x_1 + 1.5x_2 \leq 3000$

Additionally, since we cannot produce a negative number of batches, we have non-negativity constraints:

- $x_1 \geq 0$
- $x_2 \geq 0$

Thus, the symbolic representation of the problem in JSON format is:
```json
{
    'sym_variables': [('x1', 'number of batches of sausages'), ('x2', 'number of batches of burger patties')],
    'objective_function': '200*x1 + 250*x2',
    'constraints': ['2*x1 + 4*x2 <= 3000', '3*x1 + 1.5*x2 <= 3000', 'x1 >= 0', 'x2 >= 0']
}
```

Now, let's implement the solution using Gurobi in Python:
```python
from gurobipy import *

# Create a new model
model = Model("Sausage_and_Burger_Patties")

# Define variables
x1 = model.addVar(vtype=GRB.CONTINUOUS, name="sausages", lb=0)
x2 = model.addVar(vtype=GRB.CONTINUOUS, name="burger_patties", lb=0)

# Set the objective function
model.setObjective(200*x1 + 250*x2, GRB.MAXIMIZE)

# Add constraints
model.addConstr(2*x1 + 4*x2 <= 3000, "meat_grinder")
model.addConstr(3*x1 + 1.5*x2 <= 3000, "meat_packer")

# Optimize the model
model.optimize()

# Print the solution
if model.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Number of batches of sausages: {x1.x}")
    print(f"Number of batches of burger patties: {x2.x}")
    print(f"Maximum profit: {model.objVal}")
else:
    print("No optimal solution found")
```