## Step 1: Define the symbolic representation of the problem
Let's denote the number of servings of cereal A as $x_1$ and the number of servings of cereal B as $x_2$. The objective is to minimize the cost, which is $0.45x_1 + 0.55x_2$. The constraints are:
- $25x_1 + 20x_2 \geq 400$ (at least 400 units of iron)
- $30x_1 + 40x_2 \geq 450$ (at least 450 units of fiber)
- $x_1 \geq 0$ and $x_2 \geq 0$ (non-negativity constraints, as the number of servings cannot be negative)

## Step 2: Express the problem in the required symbolic format
The symbolic variables are:
- $x_1$ for cereal A servings
- $x_2$ for cereal B servings

The objective function to minimize is: $0.45x_1 + 0.55x_2$

The constraints are:
- $25x_1 + 20x_2 \geq 400$
- $30x_1 + 40x_2 \geq 450$
- $x_1 \geq 0$
- $x_2 \geq 0$

In the required format:
```json
{
'sym_variables': [('x1', 'cereal A servings'), ('x2', 'cereal B servings')],
'objective_function': '0.45*x1 + 0.55*x2',
'constraints': ['25*x1 + 20*x2 >= 400', '30*x1 + 40*x2 >= 450', 'x1 >= 0', 'x2 >= 0']
}
```

## Step 3: Convert the problem into Gurobi code
To solve this linear programming problem using Gurobi, we will use the Gurobi Python API.

```python
import gurobi

def solve_cereal_problem():
    # Create a new model
    model = gurobi.Model()

    # Define the variables
    x1 = model.addVar(name="cereal_A", lb=0, vtype=gurobi.GRB.CONTINUOUS)
    x2 = model.addVar(name="cereal_B", lb=0, vtype=gurobi.GRB.CONTINUOUS)

    # Set the objective function
    model.setObjective(0.45 * x1 + 0.55 * x2, gurobi.GRB.MINIMIZE)

    # Add constraints
    model.addConstr(25 * x1 + 20 * x2 >= 400, name="iron_constraint")
    model.addConstr(30 * x1 + 40 * x2 >= 450, name="fiber_constraint")

    # Optimize the model
    model.optimize()

    # Print the solution
    if model.status == gurobi.GRB.OPTIMAL:
        print("Optimal solution found.")
        print(f"Cereal A servings: {x1.varValue}")
        print(f"Cereal B servings: {x2.varValue}")
        print(f"Total cost: {model.objVal}")
    else:
        print("No optimal solution found.")

solve_cereal_problem()
```