To solve this optimization problem, we first need to translate the given natural language description into a symbolic representation. This involves defining variables for each item (bowls of pasta and corn cobs), expressing the objective function using these variables, and listing all constraints in terms of these variables.

Let's denote:
- \(x_1\) as the number of bowls of pasta,
- \(x_2\) as the number of corn cobs.

The objective function to minimize is given by: 
\[6.74x_1 + 9.12x_2\]

Constraints are as follows:
1. Carbohydrates from bowls of pasta and corn cobs should be at least 16 grams: \(11x_1 + 10x_2 \geq 16\)
2. The same constraint is repeated, so we consider it only once.
3. Constraint involving the number of bowls of pasta and corn cobs: \(8x_1 - 6x_2 \geq 0\)
4. Total carbohydrates from both sources should not exceed 50 grams: \(11x_1 + 10x_2 \leq 50\)
5. Both \(x_1\) and \(x_2\) must be whole numbers.

Symbolic representation:
```json
{
    'sym_variables': [('x1', 'bowls of pasta'), ('x2', 'corn cobs')],
    'objective_function': '6.74*x1 + 9.12*x2',
    'constraints': [
        '11*x1 + 10*x2 >= 16',
        '8*x1 - 6*x2 >= 0',
        '11*x1 + 10*x2 <= 50',
        'x1 >= 0', 
        'x2 >= 0',
        'x1 == int(x1)',
        'x2 == int(x2)'
    ]
}
```

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

```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(vtype=GRB.INTEGER, name="bowls_of_pasta")
x2 = m.addVar(vtype=GRB.INTEGER, name="corn_cobs")

# Objective function
m.setObjective(6.74*x1 + 9.12*x2, GRB.MINIMIZE)

# Constraints
m.addConstr(11*x1 + 10*x2 >= 16, "carbohydrates_min")
m.addConstr(8*x1 - 6*x2 >= 0, "pasta_corn_relation")
m.addConstr(11*x1 + 10*x2 <= 50, "total_carbohydrates_max")

# Non-negativity constraints (implied by GRB.INTEGER and default lower bound of 0)

# Solve the model
m.optimize()

# Print solution
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Number of bowls of pasta: {x1.x}")
    print(f"Number of corn cobs: {x2.x}")
    print(f"Objective function value: {m.objVal}")
else:
    print("No optimal solution found")
```