To solve the given optimization problem using Gurobi, we first need to understand and possibly simplify or restate the problem in a more mathematical form that can be directly translated into code.

Given:
- Variables: `x0` (strips of bacon), `x1` (oranges)
- Objective Function: Maximize `1.28*x0^2 + 3.91*x1^2 + 8.56*x0 + 5.2*x1`
- Constraints:
  1. Fiber from strips of bacon and oranges: `8*x0 + 3*x1 >= 14`
  2. Four times the number of strips of bacon minus six times the number of oranges must be at least zero: `4*x0 - 6*x1 >= 0`
  3. Up to 15 grams of fiber from squares: `(8*x0^2 + 3*x1^2) <= 15` (Note: This constraint seems unusual because it mixes linear and quadratic terms in a way that doesn't directly relate to typical fiber calculations, but we'll implement as stated)
  4. At most 15 grams of fiber from strips of bacon plus oranges: `8*x0 + 3*x1 <= 15`
  5. Integer number of strips of bacon: `x0` must be an integer
  6. Non-integer number of oranges allowed: `x1` can be any real number

We need to translate these requirements into Gurobi code, which is typically used in Python.

```python
from gurobipy import *

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

# Define variables
x0 = m.addVar(vtype=GRB.INTEGER, name="strips_of_bacon")  # Integer variable for strips of bacon
x1 = m.addVar(vtype=GRB.CONTINUOUS, name="oranges")  # Continuous variable for oranges

# Objective function: Maximize
m.setObjective(1.28*x0**2 + 3.91*x1**2 + 8.56*x0 + 5.2*x1, GRB.MAXIMIZE)

# Constraints
m.addConstr(8*x0 + 3*x1 >= 14, name="min_fiber")  # Minimum fiber constraint
m.addConstr(4*x0 - 6*x1 >= 0, name="bacon_orange_ratio")  # Bacon to orange ratio constraint
m.addConstr((8*x0**2 + 3*x1**2) <= 15, name="max_fiber_squared")  # Maximum fiber from squares
m.addConstr(8*x0 + 3*x1 <= 15, name="max_fiber_linear")  # Maximum fiber linear constraint

# Optimize the model
m.optimize()

# Print results if optimized successfully
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Strips of bacon: {x0.X}")
    print(f"Oranges: {x1.X}")
    print(f"Objective function value: {m.ObjVal}")
else:
    print("No optimal solution found")
```