To solve the given optimization problem, we first need to understand the objective function and the constraints. The objective is to minimize the cost function: $8.4 \times \text{green beans} + 6.91 \times \text{hamburgers}$.

The constraints are as follows:
1. Each green bean contains 5 milligrams of calcium.
2. Each hamburger contains 4 milligrams of calcium.
3. The total calcium from green beans and hamburgers must be at least 10 milligrams.
4. The constraint about getting at least 10 milligrams of calcium from both green beans and hamburgers seems to be a repetition or misinterpretation since the original problem statement does not specify that each item individually must provide this amount, but rather it's the total that matters. Thus, we'll consider the combined requirement.
5. The number of green beans minus the number of hamburgers must be at least zero.
6. The total calcium from green beans and hamburgers cannot exceed 21 milligrams.

Given these conditions, we can model this problem using linear programming techniques since all constraints and the objective function are linear.

Let $x_0$ represent the amount of green beans and $x_1$ represent the number of hamburgers.

The objective function to minimize is: $8.4x_0 + 6.91x_1$

Constraints:
- Calcium from green beans and hamburgers: $5x_0 + 4x_1 \geq 10$
- The upper limit of calcium intake: $5x_0 + 4x_1 \leq 21$
- Green beans minus hamburgers: $x_0 - x_1 \geq 0$

Since we are allowed to use fractional numbers of both green beans and hamburgers, both $x_0$ and $x_1$ are continuous variables.

Here is the Gurobi code that captures this problem:

```python
from gurobipy import *

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

# Define the variables
x0 = m.addVar(lb=0, ub=GRB.INFINITY, vtype=GRB.CONTINUOUS, name="green_beans")
x1 = m.addVar(lb=0, ub=GRB.INFINITY, vtype=GRB.CONTINUOUS, name="hamburgers")

# Define the objective function
m.setObjective(8.4*x0 + 6.91*x1, GRB.MINIMIZE)

# Add constraints
m.addConstr(5*x0 + 4*x1 >= 10, "calcium_minimum")
m.addConstr(5*x0 + 4*x1 <= 21, "calcium_maximum")
m.addConstr(x0 - x1 >= 0, "green_beans_minus_hamburgers")

# Optimize the model
m.optimize()

# Print solution
if m.status == GRB.OPTIMAL:
    print("Optimal solution found.")
    print("Green Beans:", x0.x)
    print("Hamburgers:", x1.x)
    print("Objective Function Value:", m.objVal)
else:
    print("No optimal solution found. Status:", m.status)
```