To solve the given optimization problem, we first need to translate the natural language description into a symbolic representation. This involves defining variables for the quantities of platinum and silver necklaces, formulating the objective function that represents the total profit, and listing the constraints based on the availability of designing and crafting teams.

Let's define:
- \(x_1\) as the number of platinum necklaces.
- \(x_2\) as the number of silver necklaces.

The profit per platinum necklace is $2000, and per silver necklace is $700. Thus, the objective function to maximize the total profit can be written as:
\[ \text{Maximize:} \quad 2000x_1 + 700x_2 \]

The constraints are based on the time availability for designing and crafting:
- Designing team constraint: \(4x_1 + 7x_2 \leq 35\)
- Crafting team constraint: \(15x_1 + 5x_2 \leq 40\)

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

Thus, the symbolic representation of the problem in JSON format is:

```json
{
    'sym_variables': [('x1', 'platinum necklaces'), ('x2', 'silver necklaces')],
    'objective_function': '2000*x1 + 700*x2',
    'constraints': ['4*x1 + 7*x2 <= 35', '15*x1 + 5*x2 <= 40', 'x1 >= 0', 'x2 >= 0']
}
```

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

```python
from gurobipy import *

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

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

# Set the objective function
m.setObjective(2000*x1 + 700*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(4*x1 + 7*x2 <= 35, "designing_team")
m.addConstr(15*x1 + 5*x2 <= 40, "crafting_team")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Platinum necklaces: {x1.x}")
    print(f"Silver necklaces: {x2.x}")
    print(f"Total profit: ${2000*x1.x + 700*x2.x:.2f}")
else:
    print("No optimal solution found.")
```