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 define:
- $x_1$ as the number of cutting boards made,
- $x_2$ as the number of chairs made.

The objective is to maximize profit. Given that each cutting board yields a profit of $14 and each chair yields a profit of $25, the objective function can be written as:
\[ \text{Maximize:} \quad 14x_1 + 25x_2 \]

The constraints are based on the time available and the total number of items that can be made:
1. Time constraint: Each cutting board takes 30 minutes to make, and each chair takes 70 minutes to make. The artist has 1500 minutes available.
\[ 30x_1 + 70x_2 \leq 1500 \]
2. Total item constraint: The artist only has enough wood to make a total of 40 items.
\[ x_1 + x_2 \leq 40 \]
3. Non-negativity constraints: The number of cutting boards and chairs made cannot be negative.
\[ x_1 \geq 0, \quad x_2 \geq 0 \]

In symbolic representation, the problem can be described as:
```json
{
    'sym_variables': [('x1', 'number of cutting boards'), ('x2', 'number of chairs')],
    'objective_function': '14*x1 + 25*x2',
    'constraints': ['30*x1 + 70*x2 <= 1500', 'x1 + x2 <= 40', 'x1 >= 0', 'x2 >= 0']
}
```

Now, let's implement this optimization problem using Gurobi in Python:

```python
from gurobipy import *

# Create a model
m = Model("Wood_Artist_Problem")

# Define variables
x1 = m.addVar(name='cutting_boards', vtype=GRB.INTEGER, lb=0)
x2 = m.addVar(name='chairs', vtype=GRB.INTEGER, lb=0)

# Set objective function
m.setObjective(14*x1 + 25*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(30*x1 + 70*x2 <= 1500, name='time_constraint')
m.addConstr(x1 + x2 <= 40, name='total_items_constraint')

# Optimize model
m.optimize()

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