To solve the optimization problem described, we first need to convert the natural language description into a symbolic representation. This involves defining variables, an objective function, and constraints based on the given information.

Let's denote:
- $x_1$ as the number of glass chandeliers.
- $x_2$ as the number of brass chandeliers.

The profit per glass chandelier is $400, and per brass chandelier is $300. Therefore, the objective function to maximize profit can be represented as:
\[ 400x_1 + 300x_2 \]

Given the constraints:
- Each glass chandelier takes 2 hours for crafting, and each brass chandelier takes 1.5 hours for crafting. The company has available 750 hours for crafting.
- Each glass chandelier takes 1 hour for installation, and each brass chandelier takes 0.75 hours for installation. The company has available 500 hours for installation.

The constraints can be represented as:
1. $2x_1 + 1.5x_2 \leq 750$ (crafting time constraint)
2. $x_1 + 0.75x_2 \leq 500$ (installation time constraint)
3. $x_1, x_2 \geq 0$ (non-negativity constraints, as the number of chandeliers cannot be negative)

Symbolic representation in JSON format:
```json
{
    'sym_variables': [('x1', 'number of glass chandeliers'), ('x2', 'number of brass chandeliers')],
    'objective_function': '400*x1 + 300*x2',
    'constraints': ['2*x1 + 1.5*x2 <= 750', 'x1 + 0.75*x2 <= 500', 'x1 >= 0', 'x2 >= 0']
}
```

To solve this linear programming problem, we can use Gurobi in Python:

```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(lb=0, name="glass_chandeliers")
x2 = m.addVar(lb=0, name="brass_chandeliers")

# Set the objective function
m.setObjective(400*x1 + 300*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(2*x1 + 1.5*x2 <= 750, name="crafting_time")
m.addConstr(x1 + 0.75*x2 <= 500, name="installation_time")

# Optimize the model
m.optimize()

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