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

Let's define:
- $x_1$ as the number of strawberry brownies made.
- $x_2$ as the number of caramel brownies made.

The objective is to maximize revenue, given that each strawberry brownie sells for $1.5 and each caramel brownie sells for $2. Therefore, the objective function can be represented as:
\[ \text{Maximize:} \quad 1.5x_1 + 2x_2 \]

Given the constraints:
1. At least 50 strawberry brownies must be made: $x_1 \geq 50$.
2. At least 75 caramel brownies must be made: $x_2 \geq 75$.
3. No more than 100 strawberry brownies can be made: $x_1 \leq 100$.
4. No more than 150 caramel brownies can be made: $x_2 \leq 150$.
5. The bakery makes at least twice as many strawberry brownies as caramel brownies: $x_1 \geq 2x_2$.

The symbolic representation of the problem is:
```json
{
    'sym_variables': [('x1', 'strawberry brownies'), ('x2', 'caramel brownies')],
    'objective_function': '1.5*x1 + 2*x2',
    'constraints': [
        'x1 >= 50',
        'x2 >= 75',
        'x1 <= 100',
        'x2 <= 150',
        'x1 >= 2*x2'
    ]
}
```

To solve this problem using Gurobi in Python, we will use the following code:

```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(lb=0, vtype=GRB.INTEGER, name="strawberry_brownies")
x2 = m.addVar(lb=0, vtype=GRB.INTEGER, name="caramel_brownies")

# Set the objective function
m.setObjective(1.5*x1 + 2*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(x1 >= 50, "min_strawberry")
m.addConstr(x2 >= 75, "min_caramel")
m.addConstr(x1 <= 100, "max_strawberry")
m.addConstr(x2 <= 150, "max_caramel")
m.addConstr(x1 >= 2*x2, "strawberry_vs_caramel")

# Optimize model
m.optimize()

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