To solve the optimization problem described, 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 given information.

Let's denote:
- $x_1$ as the number of vegetarian meals eaten,
- $x_2$ as the number of meat meals eaten.

The objective is to minimize costs. Given that a vegetarian meal costs $4 and a meat meal costs $6, the objective function can be written as:
\[ \text{Minimize: } 4x_1 + 6x_2 \]

The daily requirements are at least 100 grams of protein and 150 grams of carbs. Since a vegetarian meal contains 10 grams of protein and 20 grams of carbs, and a meat meal contains 30 grams of protein and 15 grams of carbs, we can write the constraints as follows:
- Protein constraint: $10x_1 + 30x_2 \geq 100$
- Carbs constraint: $20x_1 + 15x_2 \geq 150$

Additionally, since the number of meals cannot be negative, we have non-negativity constraints:
- $x_1 \geq 0$
- $x_2 \geq 0$

Now, let's represent this problem symbolically as requested:

```json
{
    'sym_variables': [('x1', 'number of vegetarian meals'), ('x2', 'number of meat meals')],
    'objective_function': '4*x1 + 6*x2',
    'constraints': ['10*x1 + 30*x2 >= 100', '20*x1 + 15*x2 >= 150', 'x1 >= 0', 'x2 >= 0']
}
```

Next, we'll translate this symbolic representation into Gurobi code in Python:

```python
from gurobipy import *

# Create a model
m = Model("Meal_Optimization")

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

# Set the objective function
m.setObjective(4*x1 + 6*x2, GRB.MINIMIZE)

# Add constraints
m.addConstr(10*x1 + 30*x2 >= 100, "protein_constraint")
m.addConstr(20*x1 + 15*x2 >= 150, "carbs_constraint")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Number of vegetarian meals: {x1.x}")
    print(f"Number of meat meals: {x2.x}")
    print(f"Total cost: ${4*x1.x + 6*x2.x:.2f}")
else:
    print("No optimal solution found")
```