To solve the given problem, we first need to convert the natural language description into a symbolic representation. This involves defining variables for the quantities of bracelets and rings to be made, formulating the objective function that represents the total profit, and listing all the constraints based on the production limits and approval capacity.

Let's define:
- $x_1$ as the number of bracelets to be made,
- $x_2$ as the number of rings to be made.

The profit per bracelet is $700, and the profit per ring is $300. Thus, the objective function to maximize profit can be written as:
\[ 700x_1 + 300x_2 \]

The constraints are:
1. The team making bracelets can make at most 4 bracelets per day: $x_1 \leq 4$
2. The team making rings can make at most 7 rings per day: $x_2 \leq 7$
3. The master jeweler can check at most 30 jewels (bracelets or rings) per day: $x_1 + x_2 \leq 30$

Also, since the number of bracelets and rings cannot be negative:
- $x_1 \geq 0$
- $x_2 \geq 0$

Symbolic representation in JSON format:
```json
{
    'sym_variables': [('x1', 'number of bracelets'), ('x2', 'number of rings')],
    'objective_function': '700*x1 + 300*x2',
    'constraints': ['x1 <= 4', 'x2 <= 7', 'x1 + x2 <= 30', 'x1 >= 0', 'x2 >= 0']
}
```

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

```python
from gurobipy import *

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

# Define the variables
x1 = m.addVar(lb=0, vtype=GRB.INTEGER, name="number_of_bracelets")
x2 = m.addVar(lb=0, vtype=GRB.INTEGER, name="number_of_rings")

# Set the objective function
m.setObjective(700*x1 + 300*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(x1 <= 4, "bracelet_production_limit")
m.addConstr(x2 <= 7, "ring_production_limit")
m.addConstr(x1 + x2 <= 30, "jeweler_approval_limit")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Number of bracelets: {x1.x}")
    print(f"Number of rings: {x2.x}")
    print(f"Maximum profit: {m.objVal}")
else:
    print("No optimal solution found")
```