To solve this problem, we first need to define the decision variables and the objective function. Let's denote the number of batches of bagels as \(B\) and the number of batches of croissants as \(C\). The profit per batch of bagels is $7.5, and the profit per batch of croissants is $5. Thus, the total profit \(P\) can be represented as:

\[ P = 7.5B + 5C \]

We need to maximize \(P\), subject to the constraints imposed by the availability of the dough mixer and the commercial bake oven.

1. **Dough Mixer Constraint**: Each batch of bagels requires 2 hours in the dough mixer, and each batch of croissants requires 1.5 hours. Given that the dough mixer can run for at most 2500 hours per year, we have:

\[ 2B + 1.5C \leq 2500 \]

2. **Oven Constraint**: Each batch of bagels requires 3.5 hours in the oven, and each batch of croissants requires 2 hours. Given that the oven can also run for at most 2500 hours per year, we have:

\[ 3.5B + 2C \leq 2500 \]

3. **Non-Negativity Constraints**: Both \(B\) and \(C\) must be non-negative since they represent the number of batches.

Now, let's translate these constraints and the objective function into Gurobi code in Python:

```python
from gurobipy import *

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

# Define the decision variables
B = m.addVar(name="Bagels", vtype=GRB.CONTINUOUS, lb=0)
C = m.addVar(name="Croissants", vtype=GRB.CONTINUOUS, lb=0)

# Set the objective function: Maximize profit
m.setObjective(7.5*B + 5*C, GRB.MAXIMIZE)

# Add constraints
m.addConstr(2*B + 1.5*C <= 2500, name="Dough_Mixer_Constraint")
m.addConstr(3.5*B + 2*C <= 2500, name="Oven_Constraint")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Bagels: {B.x}")
    print(f"Croissants: {C.x}")
    print(f"Total Profit: ${7.5*B.x + 5*C.x:.2f}")
else:
    print("No optimal solution found.")
```