To solve the optimization problem described, we first need to translate the natural language description into a symbolic representation. This involves defining variables for the quantities of Pizza A and Pizza B to be produced, formulating the objective function that represents the total profit, and listing the constraints based on the availability of mozzarella and American cheese.

Let's define:
- $x_1$ as the number of Pizza A to be made,
- $x_2$ as the number of Pizza B to be made.

The objective function aims to maximize the total profit. Given that the profit per pizza A is $3 and per pizza B is $4, the objective function can be written as:
\[ \text{Maximize:} \quad 3x_1 + 4x_2 \]

The constraints are based on the availability of cheese:
- Mozzarella cheese constraint: $4x_1 + 5x_2 \leq 600$ (since Pizza A requires 4 units and Pizza B requires 5 units, and there are 600 units available),
- American cheese constraint: $5x_1 + 3x_2 \leq 500$ (since Pizza A requires 5 units and Pizza B requires 3 units, and there are 500 units available),
- Non-negativity constraints: $x_1 \geq 0$ and $x_2 \geq 0$, since the number of pizzas cannot be negative.

Thus, the symbolic representation of the problem is:
```json
{
    'sym_variables': [('x1', 'Number of Pizza A'), ('x2', 'Number of Pizza B')],
    'objective_function': '3*x1 + 4*x2',
    'constraints': ['4*x1 + 5*x2 <= 600', '5*x1 + 3*x2 <= 500', 'x1 >= 0', 'x2 >= 0']
}
```

Now, let's implement this problem in Gurobi using Python:

```python
from gurobipy import *

# Create a new model
model = Model("Pizza_Optimization")

# Define variables
x1 = model.addVar(vtype=GRB.CONTINUOUS, name="Pizza_A", lb=0)
x2 = model.addVar(vtype=GRB.CONTINUOUS, name="Pizza_B", lb=0)

# Set the objective function
model.setObjective(3*x1 + 4*x2, GRB.MAXIMIZE)

# Add constraints
model.addConstr(4*x1 + 5*x2 <= 600, "Mozzarella_Cheese_Constraint")
model.addConstr(5*x1 + 3*x2 <= 500, "American_Cheese_Constraint")

# Optimize the model
model.optimize()

# Print the results
if model.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Pizza A: {x1.x}")
    print(f"Pizza B: {x2.x}")
    print(f"Total Profit: {model.ObjVal}")
else:
    print("No optimal solution found")
```