To solve the optimization problem described, we first need to convert the natural language description into a symbolic representation. This involves defining variables, an objective function, and constraints based on the given information.

Let's denote:
- \(x_1\) as the number of AA batteries to stock,
- \(x_2\) as the number of D batteries to stock.

The objective is to maximize profit. Given that the profit per AA battery is $0.50 and the profit per D battery is $1, the objective function can be represented as:
\[ \text{Maximize: } 0.5x_1 + x_2 \]

The constraints are based on the budget and the demand:
1. Budget constraint: The total cost of AA and D batteries must not exceed $1000. Given that each AA battery costs $1 and each D battery costs $3, we have:
\[ x_1 + 3x_2 \leq 1000 \]
2. Demand constraint: The monthly demand for both batteries will not exceed 1000, so we have:
\[ x_1 + x_2 \leq 1000 \]
3. Non-negativity constraints: The number of batteries cannot be negative, so we also have:
\[ x_1 \geq 0 \]
\[ x_2 \geq 0 \]

Thus, the symbolic representation of the problem is:
```json
{
    'sym_variables': [('x1', 'Number of AA batteries'), ('x2', 'Number of D batteries')],
    'objective_function': 'Maximize: 0.5*x1 + x2',
    'constraints': ['x1 + 3*x2 <= 1000', 'x1 + x2 <= 1000', 'x1 >= 0', 'x2 >= 0']
}
```

Now, let's implement this problem using Gurobi in Python:

```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(name="AA_Batteries", lb=0)  # Number of AA batteries
x2 = m.addVar(name="D_Batteries", lb=0)   # Number of D batteries

# Set the objective function
m.setObjective(0.5*x1 + x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(x1 + 3*x2 <= 1000, name="Budget_Constraint")
m.addConstr(x1 + x2 <= 1000, name="Demand_Constraint")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Number of AA batteries: {x1.x}")
    print(f"Number of D batteries: {x2.x}")
    print(f"Maximum Profit: {m.objVal}")
else:
    print("No optimal solution found")
```