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

Let's define:
- $x_1$ as the number of tons of cocoa beans produced per day.
- $x_2$ as the number of tons of coffee beans produced per day.

The objective is to maximize profit. Given that the profit per ton of cocoa beans is $500 and the profit per ton of coffee beans is $750, the objective function can be written as:
\[ \text{Maximize: } 500x_1 + 750x_2 \]

Now, let's consider the constraints:
1. Production capacity constraint: The factory has a production capacity of 15 tons per day for both cocoa and coffee beans combined.
\[ x_1 + x_2 \leq 15 \]
2. Roasting machine availability constraint: Each ton requires 15 hours of roasting, and the roasting machine is available for at most 1000 hours.
\[ 15x_1 + 15x_2 \leq 1000 \]
3. Minimum production requirements:
- For cocoa beans: $x_1 \geq 3$
- For coffee beans: $x_2 \geq 5$

Thus, the symbolic representation of the problem can be summarized as follows:

```json
{
    'sym_variables': [('x1', 'tons of cocoa beans'), ('x2', 'tons of coffee beans')],
    'objective_function': '500*x1 + 750*x2',
    'constraints': [
        'x1 + x2 <= 15',
        '15*x1 + 15*x2 <= 1000',
        'x1 >= 3',
        'x2 >= 5'
    ]
}
```

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

```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(lb=3, name="cocoa_beans")  # Tons of cocoa beans
x2 = m.addVar(lb=5, name="coffee_beans")  # Tons of coffee beans

# Set the objective function
m.setObjective(500*x1 + 750*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(x1 + x2 <= 15, "production_capacity")
m.addConstr(15*x1 + 15*x2 <= 1000, "roasting_machine_availability")

# Optimize model
m.optimize()

# Print solution
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Cocoa beans: {x1.x} tons")
    print(f"Coffee beans: {x2.x} tons")
    print(f"Total profit: ${500*x1.x + 750*x2.x}")
else:
    print("No optimal solution found")
```