To solve the given optimization problem, 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 information provided.

Let's define:
- \(x_1\) as the number of small puzzles made,
- \(x_2\) as the number of large puzzles made.

The objective is to maximize profit. Given that the profit per small puzzle is $8 and the profit per large puzzle is $12, the objective function can be written as:
\[ \text{Maximize: } 8x_1 + 12x_2 \]

The constraints are based on the time available for cutting and quality checking:
- Each small puzzle takes 10 minutes of cutting, and each large puzzle takes 15 minutes. The company has 600 minutes available for cutting, so we have:
\[ 10x_1 + 15x_2 \leq 600 \]
- Each small puzzle takes 20 minutes of quality checking, and each large puzzle takes 30 minutes. With 1000 minutes available for quality checking, we get:
\[ 20x_1 + 30x_2 \leq 1000 \]

Additionally, \(x_1\) and \(x_2\) must be non-negative since they represent the number of puzzles.

Thus, the symbolic representation of the problem is:
```json
{
  'sym_variables': [('x1', 'number of small puzzles'), ('x2', 'number of large puzzles')],
  'objective_function': '8*x1 + 12*x2',
  'constraints': ['10*x1 + 15*x2 <= 600', '20*x1 + 30*x2 <= 1000', 'x1 >= 0', 'x2 >= 0']
}
```

Now, let's write the Gurobi code in Python to solve this linear programming problem:

```python
from gurobipy import *

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

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

# Set the objective function
m.setObjective(8*x1 + 12*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(10*x1 + 15*x2 <= 600, "cutting_time")
m.addConstr(20*x1 + 30*x2 <= 1000, "quality_checking_time")

# Optimize the model
m.optimize()

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