To solve the given optimization problem, we first need to establish a symbolic representation of the variables and the constraints involved. The problem involves maximizing an objective function that is dependent on two variables: 'bowls of cereal' and 'pickles'. These can be represented symbolically as \(x_1\) for 'bowls of cereal' and \(x_2\) for 'pickles'.

The objective function to maximize is given by \(1.79x_1 + 4.7x_2\).

The constraints are as follows:
1. Carbohydrates from bowls of cereal: \(3x_1\)
2. Sourness index from bowls of cereal: \(6x_1\)
3. Carbohydrates from pickles: \(3x_2\)
4. Sourness index from pickles: \(2x_2\)
5. Minimum carbohydrates: \(3x_1 + 3x_2 \geq 37\)
6. Minimum sourness index: \(6x_1 + 2x_2 \geq 11\)
7. Mixed constraint: \(10x_1 - 6x_2 \geq 0\)
8. Maximum carbohydrates: \(3x_1 + 3x_2 \leq 46\)
9. Maximum sourness index: \(6x_1 + 2x_2 \leq 39\)

Given these constraints and the objective function, we can represent the problem symbolically as:

```json
{
  'sym_variables': [('x1', 'bowls of cereal'), ('x2', 'pickles')],
  'objective_function': 'Maximize 1.79*x1 + 4.7*x2',
  'constraints': [
    '3*x1 + 3*x2 >= 37',
    '6*x1 + 2*x2 >= 11',
    '10*x1 - 6*x2 >= 0',
    '3*x1 + 3*x2 <= 46',
    '6*x1 + 2*x2 <= 39'
  ]
}
```

To solve this problem using Gurobi, we can use the following Python code:

```python
from gurobipy import *

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

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

# Set the objective function
m.setObjective(1.79*x1 + 4.7*x2, GRB.MAXIMIZE)

# Add constraints to the model
m.addConstr(3*x1 + 3*x2 >= 37, "min_carbohydrates")
m.addConstr(6*x1 + 2*x2 >= 11, "min_sourness_index")
m.addConstr(10*x1 - 6*x2 >= 0, "mixed_constraint")
m.addConstr(3*x1 + 3*x2 <= 46, "max_carbohydrates")
m.addConstr(6*x1 + 2*x2 <= 39, "max_sourness_index")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"bowls_of_cereal: {x1.x}")
    print(f"pickles: {x2.x}")
    print(f"Objective function value: {m.objVal}")
else:
    print("No optimal solution found.")
```