To convert the given natural language description of an optimization problem into functional Gurobi code, we need to break down the problem into its key components: variables, objective function, and constraints.

1. **Variables**: The problem involves two variables - `knishes` and `protein_bars`. Since `knishes` must be an integer amount and `protein_bars` can be any real number, in Gurobi, we will declare `knishes` as an integer variable and `protein_bars` as a continuous variable.

2. **Objective Function**: The objective is to maximize the function \(1.71 \times (\text{knishes})^2 + 1.41 \times \text{knishes} \times \text{protein bars} + 6.44 \times (\text{protein bars})^2\).

3. **Constraints**:
   - Each knish contains 5 milligrams of iron, and each protein bar contains 5 milligrams of iron.
   - At least 46 milligrams of iron must come from the square of `knishes` and `protein_bars`.
   - \(6 \times \text{knishes} - 3 \times \text{protein bars} \geq 0\).
   - The total iron from the squares of `knishes` and `protein_bars` must not exceed 52 milligrams.
   - The total iron from `knishes` and `protein bars` must not exceed 52 milligrams.

Given these details, we can now formulate the Gurobi code to solve this optimization problem.

```python
from gurobipy import *

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

# Define variables
knishes = m.addVar(vtype=GRB.INTEGER, name="knishes")
protein_bars = m.addVar(vtype=GRB.CONTINUOUS, name="protein_bars")

# Objective function: Maximize
m.setObjective(1.71 * knishes**2 + 1.41 * knishes * protein_bars + 6.44 * protein_bars**2, GRB.MAXIMIZE)

# Constraints
# At least 46 milligrams of iron from squares
m.addConstr(knishes**2 * 5 + protein_bars**2 * 5 >= 46, name="iron_from_squares_min")

# Total iron from squares does not exceed 52 milligrams
m.addConstr(knishes**2 * 5 + protein_bars**2 * 5 <= 52, name="iron_from_squares_max")

# 6 times knishes minus 3 times protein bars is at least 0
m.addConstr(6 * knishes - 3 * protein_bars >= 0, name="knishes_protein_bars_ratio")

# Total iron from knishes and protein bars does not exceed 52 milligrams
m.addConstr(knishes * 5 + protein_bars * 5 <= 52, name="total_iron_max")

# Optimize model
m.optimize()

# Print solution
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Knishes: {knishes.x}")
    print(f"Protein Bars: {protein_bars.x}")
    print(f"Objective Function Value: {m.objVal}")
else:
    print("No optimal solution found")
```