To solve the optimization problem described, we first need to establish a clear symbolic representation of the problem. This involves defining variables for each item (slices of pizza and chicken breasts), formulating the objective function using these variables, and then translating all given constraints into algebraic expressions involving these variables.

Let's denote:
- \(x_1\) as the quantity of slices of pizza,
- \(x_2\) as the quantity of chicken breasts.

The objective function to minimize is: \(4.05x_1 + 6.13x_2\).

Given constraints are:
1. Slices of pizza cost $18 each, and chicken breasts cost $15 each.
2. The total cost must be at least $101.
3. The total cost must not exceed $177.
4. \(8x_1 - 4x_2 \geq 0\).

These constraints can be algebraically represented as:
- \(18x_1 + 15x_2 \geq 101\) (minimum cost constraint),
- \(18x_1 + 15x_2 \leq 177\) (maximum cost constraint),
- \(8x_1 - 4x_2 \geq 0\) (additional linear constraint).

The symbolic representation of the problem is:
```json
{
    'sym_variables': [('x1', 'slices of pizza'), ('x2', 'chicken breasts')],
    'objective_function': '4.05*x1 + 6.13*x2',
    'constraints': [
        '18*x1 + 15*x2 >= 101',
        '18*x1 + 15*x2 <= 177',
        '8*x1 - 4*x2 >= 0'
    ]
}
```

Now, let's implement this problem in Gurobi using Python. We'll define the model, add variables, set the objective function, and then add all constraints before solving the model.

```python
from gurobipy import *

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

# Add variables to the model
x1 = m.addVar(lb=0, name="slices_of_pizza")
x2 = m.addVar(lb=0, name="chicken_breasts")

# Set the objective function
m.setObjective(4.05*x1 + 6.13*x2, GRB.MINIMIZE)

# Add constraints to the model
m.addConstr(18*x1 + 15*x2 >= 101, "minimum_cost")
m.addConstr(18*x1 + 15*x2 <= 177, "maximum_cost")
m.addConstr(8*x1 - 4*x2 >= 0, "additional_constraint")

# Optimize the model
m.optimize()

# Print solution
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"slices_of_pizza: {x1.x}")
    print(f"chicken_breasts: {x2.x}")
    print(f"Objective Function Value: {m.objVal}")
else:
    print("No optimal solution found.")
```