To solve this linear programming optimization problem, we first need to define the decision variables and the objective function. Let's denote:

- $x$ as the number of packages of adhesives produced daily.
- $y$ as the number of packages of plasticizers produced daily.

The objective is to maximize revenue. Given that each package of adhesives sells for $8.5 and each package of plasticizers sells for $11.5, the total revenue can be represented by the equation:

\[ \text{Revenue} = 8.5x + 11.5y \]

We need to maximize this function subject to several constraints based on the availability of the devices and the production requirements.

1. **Constraint for the automatic device:** Each package of adhesives requires 6 minutes, and each package of plasticizers requires 8 minutes on the automatic device. Given that the device is available for at most 450 minutes per day, we have:

\[ 6x + 8y \leq 450 \]

2. **Constraint for the human-operated device:** Each package of adhesives requires 5 minutes, and each package of plasticizers requires 4 minutes on the human-operated device. With a maximum availability of 450 minutes per day, we get:

\[ 5x + 4y \leq 450 \]

3. **Non-negativity constraints:** Since the number of packages produced cannot be negative, we have:

\[ x \geq 0 \]
\[ y \geq 0 \]

Now, let's translate these equations into Gurobi code in Python to find out how many packages of each product should be produced daily to maximize revenue.

```python
from gurobipy import *

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

# Define the decision variables
x = m.addVar(vtype=GRB.CONTINUOUS, name="adhesives")
y = m.addVar(vtype=GRB.CONTINUOUS, name="plasticizers")

# Objective function: Maximize revenue
m.setObjective(8.5*x + 11.5*y, GRB.MAXIMIZE)

# Constraints
m.addConstr(6*x + 8*y <= 450, "automatic_device")
m.addConstr(5*x + 4*y <= 450, "human_operated_device")

# Non-negativity constraints are implicitly handled by the variable type

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Adhesives: {x.x}")
    print(f"Plasticizers: {y.x}")
    print(f"Maximum Revenue: ${8.5*x.x + 11.5*y.x:.2f}")
else:
    print("No optimal solution found")
```