## Step 1: Define the symbolic variables
Let's define the symbolic variables for the problem:
- $x_1$ represents the number of wallets.
- $x_2$ represents the number of purses.

## Step 2: Translate the objective function into symbolic notation
The profit per wallet is $40, and the profit per purse is $85. The objective function to maximize profit is:
\[ \text{Maximize:} \quad 40x_1 + 85x_2 \]

## 3: Translate the constraints into symbolic notation
The constraints based on the problem description are:
1. Leather constraint: $2x_1 + 4x_2 \leq 2000$
2. Popularity constraint: $x_2 \geq 4x_1$
3. Minimum wallets constraint: $x_1 \geq 10$
4. Non-negativity constraints: $x_1 \geq 0, x_2 \geq 0$

However, since $x_1$ and $x_2$ represent quantities of items, we can reasonably assume $x_1 \geq 0$ and $x_2 \geq 0$ are implicitly satisfied by the other constraints.

## 4: Express the problem in the required JSON format
The symbolic representation of the problem is:
```json
{
    'sym_variables': [('x1', 'wallets'), ('x2', 'purses')],
    'objective_function': '40*x1 + 85*x2',
    'constraints': [
        '2*x1 + 4*x2 <= 2000',
        'x2 >= 4*x1',
        'x1 >= 10'
    ]
}
```

## 5: Implement the problem in Gurobi code
Now, let's implement this problem using Gurobi in Python:
```python
import gurobi

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

    # Define the variables
    x1 = model.addVar(name="wallets", lb=0, vtype=gurobi.GRB.INTEGER)
    x2 = model.addVar(name="purses", lb=0, vtype=gurobi.GRB.INTEGER)

    # Objective function: Maximize 40*x1 + 85*x2
    model.setObjective(40*x1 + 85*x2, gurobi.GRB.MAXIMIZE)

    # Constraints
    model.addConstr(2*x1 + 4*x2 <= 2000, name="leather_constraint")
    model.addConstr(x2 >= 4*x1, name="popularity_constraint")
    model.addConstr(x1 >= 10, name="min_wallets_constraint")

    # Optimize the model
    model.optimize()

    # Print the solution
    if model.status == gurobi.GRB.OPTIMAL:
        print(f"Optimal solution: wallets = {x1.varValue}, purses = {x2.varValue}")
    else:
        print("No optimal solution found")

solve_leatherworker_problem()
```