## Symbolic Representation

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

The objective is to maximize profit. The profit per large bottle is $5, and per small bottle is $3. Therefore, the objective function can be represented as:
\[ \text{Maximize:} \quad 5x_1 + 3x_2 \]

The store has a budget of $1000, with each large bottle costing $3 and each small bottle costing $2. This gives us the budget constraint:
\[ 3x_1 + 2x_2 \leq 1000 \]

Each large bottle takes 2 units of shelf space, and each small bottle takes 1 unit. With 500 units of shelf space available, we have:
\[ 2x_1 + x_2 \leq 500 \]

The store wants at least 50% of all stock to be small bottles. This can be represented as:
\[ x_2 \geq 0.5(x_1 + x_2) \]
\[ 0.5x_1 - 0.5x_2 \leq 0 \]
\[ x_1 - x_2 \leq 0 \]
Or equivalently:
\[ x_2 \geq x_1 \]

Also, \(x_1 \geq 0\) and \(x_2 \geq 0\) because the number of bottles cannot be negative.

## Symbolic Representation in JSON Format
```json
{
    'sym_variables': [('x1', 'large bottles'), ('x2', 'small bottles')],
    'objective_function': '5*x1 + 3*x2',
    'constraints': [
        '3*x1 + 2*x2 <= 1000',
        '2*x1 + x2 <= 500',
        'x1 - x2 <= 0',
        'x1 >= 0',
        'x2 >= 0'
    ]
}
```

## Gurobi Code in Python
```python
import gurobi

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

    # Define variables
    x1 = model.addVar(name="large_bottles", lb=0, vtype=gurobi.GRB.INTEGER)
    x2 = model.addVar(name="small_bottles", lb=0, vtype=gurobi.GRB.INTEGER)

    # Objective function: Maximize 5*x1 + 3*x2
    model.setObjective(5*x1 + 3*x2, gurobi.GRB.MAXIMIZE)

    # Constraints
    model.addConstr(3*x1 + 2*x2 <= 1000, name="budget_constraint")
    model.addConstr(2*x1 + x2 <= 500, name="shelf_space_constraint")
    model.addConstr(x1 - x2 <= 0, name="small_bottles_proportion_constraint")

    # Solve the model
    model.optimize()

    # Check if the model is optimized
    if model.status == gurobi.GRB.OPTIMAL:
        print(f"Optimal solution: large bottles = {x1.varValue}, small bottles = {x2.varValue}")
    else:
        print("The model is infeasible")

solve_optimization_problem()
```