To solve the given optimization problem, we first need to translate the natural language description into a symbolic representation. This involves defining variables and formulating the objective function and constraints using these variables.

Let's define:
- $x_1$ as the number of bags of Bag A used in the mixture.
- $x_2$ as the number of bags of Bag B used in the mixture.

The objective is to minimize the total cost of the mixture. The cost of Bag A is $3.50 per bag, and the cost of Bag B is $2.50 per bag. Therefore, the objective function can be represented as:
\[ \text{Minimize:} \quad 3.5x_1 + 2.5x_2 \]

The constraints are as follows:
1. The mixture must contain at least 30 units of protein. Given that Bag A contains 3 units of protein per bag and Bag B contains 4 units of protein per bag, we have:
\[ 3x_1 + 4x_2 \geq 30 \]
2. The mixture must contain at least 35 units of calcium. Since Bag A contains 4 units of calcium per bag and Bag B contains 2 units of calcium per bag, we have:
\[ 4x_1 + 2x_2 \geq 35 \]

Additionally, $x_1$ and $x_2$ should be non-negative since they represent the number of bags.

Therefore, the symbolic representation of the problem is:
```json
{
  'sym_variables': [('x1', 'Number of bags of Bag A'), ('x2', 'Number of bags of Bag B')],
  'objective_function': '3.5*x1 + 2.5*x2',
  'constraints': ['3*x1 + 4*x2 >= 30', '4*x1 + 2*x2 >= 35', 'x1 >= 0', 'x2 >= 0']
}
```

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

```python
from gurobipy import *

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

# Define the variables
x1 = m.addVar(vtype=GRB.CONTINUOUS, name="bags_of_A", lb=0)
x2 = m.addVar(vtype=GRB.CONTINUOUS, name="bags_of_B", lb=0)

# Set the objective function
m.setObjective(3.5*x1 + 2.5*x2, GRB.MINIMIZE)

# Add constraints
m.addConstr(3*x1 + 4*x2 >= 30, "protein_constraint")
m.addConstr(4*x1 + 2*x2 >= 35, "calcium_constraint")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Number of bags of Bag A: {x1.x}")
    print(f"Number of bags of Bag B: {x2.x}")
    print(f"Total cost: {m.objVal}")
else:
    print("No optimal solution found")
```