To solve the given optimization problem, we first need to convert the natural language description into a symbolic representation. This involves defining variables, an objective function, and constraints based on the problem statement.

Let's define:
- $x_1$ as the amount of green apples used in kilograms.
- $x_2$ as the amount of red apples used in kilograms.

The objective is to minimize the cost of the apples used. The cost of green apples is $9 per kilogram, and the cost of red apples is $7 per kilogram. Therefore, the objective function can be represented as:
\[ \text{Minimize:} \quad 9x_1 + 7x_2 \]

The constraints are based on the requirements for sugar and fiber content in the pie filling:
- The minimum amount of sugar required is 12 kilograms. Green apples contain 5% sugar, and red apples contain 25% sugar. This translates to:
\[ 0.05x_1 + 0.25x_2 \geq 12 \]
- The minimum amount of fiber required is 5 kilograms. Green apples contain 16% fiber, and red apples contain 8% fiber. This gives us:
\[ 0.16x_1 + 0.08x_2 \geq 5 \]
- Additionally, the amounts of green and red apples used cannot be negative, so we have:
\[ x_1 \geq 0 \]
\[ x_2 \geq 0 \]

The symbolic representation of the problem is:
```json
{
    'sym_variables': [('x1', 'amount of green apples in kg'), ('x2', 'amount of red apples in kg')],
    'objective_function': '9*x1 + 7*x2',
    'constraints': ['0.05*x1 + 0.25*x2 >= 12', '0.16*x1 + 0.08*x2 >= 5', 'x1 >= 0', 'x2 >= 0']
}
```

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

```python
from gurobipy import *

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

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

# Set the objective function
m.setObjective(9*x1 + 7*x2, GRB.MINIMIZE)

# Add constraints
m.addConstr(0.05*x1 + 0.25*x2 >= 12, "sugar_constraint")
m.addConstr(0.16*x1 + 0.08*x2 >= 5, "fiber_constraint")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Green apples: {x1.x} kg")
    print(f"Red apples: {x2.x} kg")
    print(f"Minimum cost: ${m.objVal}")
else:
    print("No optimal solution found")

```