To solve Jane's problem using linear programming with Gurobi in Python, we first need to define the variables and constraints based on the given information.

Let's denote:
- \(x_A\) as the number of scoops of Alpha powder Jane should take.
- \(x_B\) as the number of scoops of Beta powder Jane should take.

The objective is to minimize the total cost. The cost per scoop of Alpha (\(x_A\)) is $1, and the cost per scoop of Beta (\(x_B\)) is $2. Therefore, the objective function can be written as:
\[ \text{Minimize} \quad 1x_A + 2x_B \]

Given that a scoop of Alpha contains 5 grams of iron and 20 grams of biotin, and a scoop of Beta contains 10 grams of iron and 3 grams of biotin, we have the following constraints based on Jane's daily requirements:
- For iron: \(5x_A + 10x_B \geq 50\)
- For biotin: \(20x_A + 3x_B \geq 40\)

Also, since Jane cannot take a negative number of scoops, we have non-negativity constraints:
\[ x_A \geq 0 \]
\[ x_B \geq 0 \]

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

```python
from gurobipy import *

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

# Define variables
x_A = m.addVar(lb=0, vtype=GRB.CONTINUOUS, name="Alpha_Scoops")
x_B = m.addVar(lb=0, vtype=GRB.CONTINUOUS, name="Beta_Scoops")

# Set the objective function
m.setObjective(1*x_A + 2*x_B, GRB.MINIMIZE)

# Add constraints
m.addConstr(5*x_A + 10*x_B >= 50, "Iron_Requirement")
m.addConstr(20*x_A + 3*x_B >= 40, "Biotin_Requirement")

# Optimize the model
m.optimize()

# Print the solution
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Alpha scoops: {x_A.x}")
    print(f"Beta scoops: {x_B.x}")
    print(f"Total cost: ${1*x_A.x + 2*x_B.x:.2f}")
else:
    print("No optimal solution found. The problem might be infeasible.")
```