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 based on the given information.

Let's define:
- $x_1$ as the number of chocolate bars produced,
- $x_2$ as the number of chocolate wafers produced.

The objective is to maximize profits, with each chocolate bar contributing $2 to profit and each chocolate wafer contributing $3. Thus, the objective function can be represented as:
\[ \text{Maximize: } 2x_1 + 3x_2 \]

The constraints are based on the production limits of each team and the shared packaging machine:
1. The chocolate bar team's production limit: $x_1 \leq 80$
2. The chocolate wafer team's production limit: $x_2 \leq 100$
3. The packaging machine's capacity: $x_1 + x_2 \leq 125$

Additionally, the number of chocolate bars and wafers cannot be negative:
- $x_1 \geq 0$
- $x_2 \geq 0$

Thus, the symbolic representation of the problem is:
```json
{
    'sym_variables': [('x1', 'number of chocolate bars'), ('x2', 'number of chocolate wafers')],
    'objective_function': '2*x1 + 3*x2',
    'constraints': ['x1 <= 80', 'x2 <= 100', 'x1 + x2 <= 125', 'x1 >= 0', 'x2 >= 0']
}
```

To solve this problem using Gurobi in Python, we can use the following code:
```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(lb=0, vtype=GRB.CONTINUOUS, name="chocolate_bars")
x2 = m.addVar(lb=0, vtype=GRB.CONTINUOUS, name="chocolate_wafers")

# Set the objective function
m.setObjective(2*x1 + 3*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(x1 <= 80, "chocolate_bar_limit")
m.addConstr(x2 <= 100, "chocolate_wafer_limit")
m.addConstr(x1 + x2 <= 125, "packaging_machine_limit")

# Optimize the model
m.optimize()

# Print the solution
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Chocolate bars: {x1.x}")
    print(f"Chocolate wafers: {x2.x}")
    print(f"Maximum profit: {m.objVal}")
else:
    print("No optimal solution found")
```