## Symbolic Representation

To solve the given problem, we first need to convert the natural language description into a symbolic representation. Let's define the variables and the objective function, and then list the constraints.

### Variables
- $x_1$ = Number of neon signs
- $x_2$ = Number of metal signs

### Objective Function
The objective is to maximize profit. Given that the profit per neon sign is $200 and the profit per metal sign is $100, the objective function can be written as:
\[ \text{Maximize:} \quad 200x_1 + 100x_2 \]

### Constraints
1. Crafting time constraint: Each neon sign takes 3 hours for crafting and each metal sign takes 2 hours for crafting, with 950 hours available.
\[ 3x_1 + 2x_2 \leq 950 \]
2. Installation time constraint: Each neon sign takes 2 hours for installation and each metal sign takes 1.5 hours for installation, with 400 hours available.
\[ 2x_1 + 1.5x_2 \leq 400 \]
3. Non-negativity constraint: The number of signs cannot be negative.
\[ x_1 \geq 0, x_2 \geq 0 \]

## Symbolic Representation in JSON Format

```json
{
    'sym_variables': [('x1', 'neon signs'), ('x2', 'metal signs')],
    'objective_function': '200*x1 + 100*x2',
    'constraints': [
        '3*x1 + 2*x2 <= 950',
        '2*x1 + 1.5*x2 <= 400',
        'x1 >= 0',
        'x2 >= 0'
    ]
}
```

## Gurobi Code in Python

```python
import gurobipy as gp

# Create a new model
model = gp.Model("Sign_Company_Optimization")

# Define variables
x1 = model.addVar(name="neon_signs", lb=0, vtype=gp.GRB.INTEGER)  # Number of neon signs
x2 = model.addVar(name="metal_signs", lb=0, vtype=gp.GRB.INTEGER)  # Number of metal signs

# Objective function: Maximize profit
model.setObjective(200 * x1 + 100 * x2, gp.GRB.MAXIMIZE)

# Constraints
model.addConstr(3 * x1 + 2 * x2 <= 950, name="crafting_time_constraint")
model.addConstr(2 * x1 + 1.5 * x2 <= 400, name="installation_time_constraint")

# Solve the model
model.optimize()

# Print the results
if model.status == gp.GRB.OPTIMAL:
    print(f"Optimal solution: {x1.varName} = {x1.x}, {x2.varName} = {x2.x}")
    print(f"Max Profit: ${200 * x1.x + 100 * x2.x}")
else:
    print("No optimal solution found.")
```