## Step 1: Define the symbolic representation of the problem

Let's denote:
- $x_1$ as the number of dining tables
- $x_2$ as the number of desks

The objective is to maximize profit, where each dining table yields a profit of $350 and each desk yields a profit of $400. Therefore, the objective function can be represented as:

Maximize $350x_1 + 400x_2$

## Step 2: Define the constraints based on the given resources

1. Woodworking: Each dining table requires 2 hours, and each desk requires 3 hours. There are 100 hours available.
\[2x_1 + 3x_2 \leq 100\]

2. Nails: Each dining table requires 3 boxes, and each desk requires 4 boxes. There are 75 boxes available.
\[3x_1 + 4x_2 \leq 75\]

3. Varnish: Each dining table requires 1 unit, and each desk requires 2 units. There are 80 units available.
\[x_1 + 2x_2 \leq 80\]

## 3: Non-negativity constraints

Since the number of dining tables and desks cannot be negative:
\[x_1 \geq 0\]
\[x_2 \geq 0\]

## 4: Symbolic representation in JSON format

```json
{
    'sym_variables': [('x1', 'dining tables'), ('x2', 'desks')],
    'objective_function': '350*x1 + 400*x2',
    'constraints': [
        '2*x1 + 3*x2 <= 100',
        '3*x1 + 4*x2 <= 75',
        'x1 + 2*x2 <= 80',
        'x1 >= 0',
        'x2 >= 0'
    ]
}
```

## 5: Gurobi code in Python

```python
import gurobi

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

    # Define variables
    x1 = model.addVar(name='dining_tables', lb=0, vtype=gurobi.GRB.CONTINUOUS)
    x2 = model.addVar(name='desks', lb=0, vtype=gurobi.GRB.CONTINUOUS)

    # Objective function
    model.setObjective(350 * x1 + 400 * x2, gurobi.GRB.MAXIMIZE)

    # Constraints
    model.addConstr(2 * x1 + 3 * x2 <= 100, name='woodworking_constraint')
    model.addConstr(3 * x1 + 4 * x2 <= 75, name='nails_constraint')
    model.addConstr(x1 + 2 * x2 <= 80, name='varnish_constraint')

    # Optimize
    model.optimize()

    # Print solution
    if model.status == gurobi.GRB.OPTIMAL:
        print(f"Optimal solution: {x1.varName} = {x1.x}, {x2.varName} = {x2.x}")
        print(f"Maximum profit: ${model.objVal:.2f}")
    else:
        print("No optimal solution found.")

solve_optimization_problem()
```