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

Let's define:
- $x_1$ as the number of carry-on suitcases produced per day.
- $x_2$ as the number of large suitcases produced per day.

The objective is to maximize profit. Given that each carry-on suitcase generates a profit of $100 and each large suitcase generates a profit of $150, the objective function can be written as:
\[ \text{Maximize: } 100x_1 + 150x_2 \]

There are several constraints based on the problem description:
1. The production rate for carry-on suitcases cannot exceed 15 per day.
\[ x_1 \leq 15 \]
2. The production rate for large suitcases cannot exceed 20 per day.
\[ x_2 \leq 20 \]
3. The total number of suitcases (both types) that can be processed by the sewing machine cannot exceed 25 per day.
\[ x_1 + x_2 \leq 25 \]
4. Non-negativity constraints, as the production of suitcases cannot be negative:
\[ x_1 \geq 0 \]
\[ x_2 \geq 0 \]

Thus, the symbolic representation of the problem is:

```json
{
    'sym_variables': [('x1', 'carry-on suitcases'), ('x2', 'large suitcases')],
    'objective_function': '100*x1 + 150*x2',
    'constraints': ['x1 <= 15', 'x2 <= 20', 'x1 + x2 <= 25', 'x1 >= 0', 'x2 >= 0']
}
```

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

```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(name="carry_on_suitcases", lb=0)
x2 = m.addVar(name="large_suitcases", lb=0)

# Set the objective function
m.setObjective(100*x1 + 150*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(x1 <= 15, name="carry_on_production_limit")
m.addConstr(x2 <= 20, name="large_suitcase_production_limit")
m.addConstr(x1 + x2 <= 25, name="sewing_machine_capacity")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print(f"Optimal production: {x1.varName} = {x1.x}, {x2.varName} = {x2.x}")
    print(f"Maximum profit: ${m.objVal}")
else:
    print("No optimal solution found")
```