## Step 1: Define the symbolic representation of the variables
The variables are 'bowls of cereal', 'oranges', and 'bowls of pasta', which can be represented symbolically as $x_0$, $x_1$, and $x_2$ respectively.

## Step 2: Convert the natural language objects into symbolic notation
The symbolic representation of the variables is:
- $x_0$ : bowls of cereal
- $x_1$ : oranges
- $x_2$ : bowls of pasta

## Step 3: Define the objective function in symbolic notation
The objective function to minimize is $8x_0 + 3x_1 + 2x_2$.

## 4: List the constraints in symbolic notation
The constraints are:
- $6x_0 + 3x_1 + 5x_2 \geq 21$ (at least 21 grams of carbohydrates from bowls of cereal and bowls of pasta)
- $6x_0 + 3x_1 + 5x_2 \geq 34$ (at least 34 grams of carbohydrates from bowls of cereal, oranges, and bowls of pasta, but this is the same as the previous one, so we keep only one)
- $6x_0 + 5x_2 \geq 37$ (at least 37 milligrams of calcium from bowls of cereal and bowls of pasta)
- $4x_0 + 4x_1 \geq 23$ (at least 23 milligrams of calcium from bowls of cereal and oranges)
- $4x_1 + x_2 \geq 34$ (at least 34 milligrams of calcium from oranges and bowls of pasta)
- $4x_0 + 4x_1 + x_2 \geq 35$ (at least 35 milligrams of calcium from bowls of cereal, oranges, and bowls of pasta, but this is essentially covered by previous constraints)
- $6x_0 + 5x_2 \leq 48$ (at most 48 grams of carbohydrates from bowls of cereal and bowls of pasta)
- $6x_0 + 3x_1 + 5x_2 \leq 73$ (at most 73 grams of carbohydrates from bowls of cereal, oranges, and bowls of pasta)

Correcting and simplifying the constraints based on the problem description:
- Carbohydrates: $6x_0 + 3x_1 + 5x_2 \geq 34$
- Calcium: $4x_0 + 4x_1 \geq 23$, $4x_0 + x_2 \geq 37$, $4x_1 + x_2 \geq 34$, and $4x_0 + 4x_1 + x_2 \geq 35$

## 5: Provide the symbolic representation of the problem
```json
{
    'sym_variables': [('x0', 'bowls of cereal'), ('x1', 'oranges'), ('x2', 'bowls of pasta')],
    'objective_function': '8*x0 + 3*x1 + 2*x2',
    'constraints': [
        '6*x0 + 3*x1 + 5*x2 >= 34',
        '4*x0 + 4*x1 >= 23',
        '4*x0 + x2 >= 37',
        '4*x1 + x2 >= 34',
        '4*x0 + 4*x1 + x2 >= 35',
        '6*x0 + 5*x2 <= 48',
        '6*x0 + 3*x1 + 5*x2 <= 73'
    ]
}
```

## 6: Implement the optimization problem using Gurobi
```python
import gurobi

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

    # Define the variables
    x0 = model.addVar(name="bowls_of_cereal", lb=0)
    x1 = model.addVar(name="oranges", lb=0)
    x2 = model.addVar(name="bowls_of_pasta", lb=0)

    # Define the objective function
    model.setObjective(8 * x0 + 3 * x1 + 2 * x2, gurobi.GRB.MINIMIZE)

    # Add constraints
    model.addConstr(6 * x0 + 3 * x1 + 5 * x2 >= 34, name="carbohydrates_total")
    model.addConstr(4 * x0 + 4 * x1 >= 23, name="calcium_cereal_orange")
    model.addConstr(4 * x0 + x2 >= 37, name="calcium_cereal_pasta")
    model.addConstr(4 * x1 + x2 >= 34, name="calcium_orange_pasta")
    model.addConstr(4 * x0 + 4 * x1 + x2 >= 35, name="calcium_all")
    model.addConstr(6 * x0 + 5 * x2 <= 48, name="carbohydrates_cereal_pasta")
    model.addConstr(6 * x0 + 3 * x1 + 5 * x2 <= 73, name="carbohydrates_all")

    # Optimize the model
    model.optimize()

    # Print the solution
    if model.status == gurobi.GRB.OPTIMAL:
        print("Optimal solution found.")
        print(f"bowls of cereal: {x0.varValue}")
        print(f"oranges: {x1.varValue}")
        print(f"bowls of pasta: {x2.varValue}")
        print(f"Objective: {model.objVal}")
    else:
        print("No optimal solution found.")

optimize_problem()
```