## Problem Description and Formulation

The cyclist needs to minimize his costs while ensuring he gets a minimum of 80 grams of protein, 50 grams of fat, and 100 grams of carbs. He can only eat chicken and potatoes, with their respective nutritional content and costs as follows:

- Chicken: 20g protein, 4g fat, 4g carbs, $6 each
- Potatoes: 2g protein, 3g fat, 7g carbs, $2 each

Let's denote the number of chickens as \(C\) and the number of potatoes as \(P\). The optimization problem can be formulated as:

### Objective Function
Minimize the total cost: \(6C + 2P\)

### Constraints
1. Protein: \(20C + 2P \geq 80\)
2. Fat: \(4C + 3P \geq 50\)
3. Carbs: \(4C + 7P \geq 100\)
4. Non-negativity: \(C \geq 0, P \geq 0\)
5. Integer: \(C, P\) are integers (though Gurobi can handle this, we will initially solve as if they are continuous and then consider if integer constraints are needed)

## Gurobi Code

```python
import gurobi

def solve_optimization_problem():
    # Create a new model
    model = gurobi.Model()

    # Define the variables
    C = model.addVar(name="chicken", lb=0, vtype=gurobi.GRB.CONTINUOUS)  # Number of chickens
    P = model.addVar(name="potato", lb=0, vtype=gurobi.GRB.CONTINUOUS)  # Number of potatoes

    # Objective function: Minimize cost
    model.setObjective(6*C + 2*P, gurobi.GRB.MINIMIZE)

    # Constraints
    model.addConstr(20*C + 2*P >= 80, name="protein_constraint")
    model.addConstr(4*C + 3*P >= 50, name="fat_constraint")
    model.addConstr(4*C + 7*P >= 100, name="carbs_constraint")

    # Solve the model
    model.optimize()

    # Print the solution
    if model.status == gurobi.GRB.OPTIMAL:
        print(f"Optimal cost: ${model.objVal:.2f}")
        print(f"Number of chickens: {C.varValue:.2f}")
        print(f"Number of potatoes: {P.varValue:.2f}")
    else:
        print("No optimal solution found.")

# Run the function
solve_optimization_problem()
```

This code sets up and solves the linear programming problem using Gurobi. Note that the solution might not be integers (e.g., 2.5 chickens), which could be problematic in practice. For an exact integer solution, you would need to adjust the `vtype` of `C` and `P` to `gurobi.GRB.INTEGER`, but this can significantly increase computation time for larger problems. 

If integer solutions are required, you can modify the `vtype` when adding the variables:

```python
C = model.addVar(name="chicken", lb=0, vtype=gurobi.GRB.INTEGER)
P = model.addVar(name="potato", lb=0, vtype=gurobi.GRB.INTEGER)
```