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

- \(x\) as the number of acres dedicated to planting cranberries.
- \(y\) as the number of acres dedicated to planting bilberries.

The objective is to maximize profit. The profit per acre from cranberries is $66, and from bilberries is $73. Therefore, the total profit can be represented as \(66x + 73y\).

We have two main constraints:

1. **Labor Constraint**: Each acre of cranberries requires 5 days of labor, and each acre of bilberries requires 4 days of labor. Bob has a total of 600 days worth of labor available. This constraint can be represented as \(5x + 4y \leq 600\).

2. **Watering Cost Constraint**: For each acre of cranberries, the watering cost is $25, and for each acre of bilberries, it's $30. Bob has a budget of $9000 for watering costs. This constraint can be represented as \(25x + 30y \leq 9000\).

Additionally, we have non-negativity constraints since Bob cannot plant a negative number of acres: \(x \geq 0\) and \(y \geq 0\). Also, the total land used cannot exceed 250 acres: \(x + y \leq 250\).

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

```python
from gurobipy import *

# Create a new model
m = Model("Berry Farm Optimization")

# Define decision variables
cranberries_acres = m.addVar(lb=0, name="Cranberries_Acres")
bilberries_acres = m.addVar(lb=0, name="Bilberries_Acres")

# Objective function: Maximize profit
m.setObjective(66*cranberries_acres + 73*bilberries_acres, GRB.MAXIMIZE)

# Constraint 1: Labor constraint
m.addConstr(5*cranberries_acres + 4*bilberries_acres <= 600, name="Labor_Constraint")

# Constraint 2: Watering cost constraint
m.addConstr(25*cranberries_acres + 30*bilberries_acres <= 9000, name="Watering_Cost_Constraint")

# Constraint 3: Total land constraint
m.addConstr(cranberries_acres + bilberries_acres <= 250, name="Total_Land_Constraint")

# Solve the model
m.optimize()

# Print results
if m.status == GRB.OPTIMAL:
    print(f"Optimal solution found:")
    print(f"Cranberries Acres: {cranberries_acres.x}")
    print(f"Bilberries Acres: {bilberries_acres.x}")
    print(f"Maximum Profit: ${m.objVal:.2f}")
else:
    print("No optimal solution found")
```