To solve Linda's bakery optimization problem, we first need to define the symbolic representation of the variables and the constraints involved.

Let's denote:
- $x_1$ as the number of chocolate chip cookies produced,
- $x_2$ as the number of oatmeal cookies produced.

The objective function is to maximize revenue. Given that each chocolate chip cookie can be sold for $4 and each oatmeal cookie can be sold for $3, the objective function can be represented algebraically as:
\[ 4x_1 + 3x_2 \]

Now, let's define the constraints based on the given resources:
1. **Mixing Time Constraint**: Each chocolate chip cookie requires 10 minutes of mixing, and each oatmeal cookie requires 20 minutes of mixing. There are 360 minutes available.
\[ 10x_1 + 20x_2 \leq 360 \]
Simplifying this, we get:
\[ x_1 + 2x_2 \leq 36 \]

2. **Vanilla Extract Constraint**: Each chocolate chip cookie requires 1 tablespoon of vanilla extract, and each oatmeal cookie requires 2 tablespoons of vanilla extract. There are 50 tablespoons available.
\[ x_1 + 2x_2 \leq 50 \]

3. **Non-Negativity Constraints**: Since Linda cannot produce a negative number of cookies, we have:
\[ x_1 \geq 0 \]
\[ x_2 \geq 0 \]

Thus, the symbolic representation of the problem in JSON format is:
```json
{
    'sym_variables': [('x1', 'number of chocolate chip cookies'), ('x2', 'number of oatmeal cookies')],
    'objective_function': '4*x1 + 3*x2',
    'constraints': ['x1 + 2*x2 <= 36', 'x1 + 2*x2 <= 50', 'x1 >= 0', 'x2 >= 0']
}
```

Now, let's implement this optimization problem using Gurobi in Python:
```python
from gurobipy import *

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

# Add variables
x1 = m.addVar(lb=0, name="chocolate_chip_cookies")
x2 = m.addVar(lb=0, name="oatmeal_cookies")

# Set the objective function
m.setObjective(4*x1 + 3*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(x1 + 2*x2 <= 36, "mixing_time_constraint")
m.addConstr(x1 + 2*x2 <= 50, "vanilla_extract_constraint")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Number of chocolate chip cookies: {x1.x}")
    print(f"Number of oatmeal cookies: {x2.x}")
    print(f"Maximum revenue: ${4*x1.x + 3*x2.x:.2f}")
else:
    print("No optimal solution found")
```