## Problem Description and Formulation

The problem is a linear programming optimization problem. The goal is to minimize the cost of producing two types of dish detergents, Fruity Loop and Passion Cook, while meeting the requirements for soap and citric acid.

Let's define the variables:

- $x_1$: the amount of Fruity Loop detergent to produce (in kg)
- $x_2$: the amount of Passion Cook detergent to produce (in kg)

The objective function is to minimize the total cost:

- Cost of Fruity Loop: $6x_1$
- Cost of Passion Cook: $5x_2$
- Total cost: $6x_1 + 5x_2$

The constraints are:

- Fruity Loop consists of 10% soap, so the amount of soap from Fruity Loop is $0.10x_1$
- Passion Cook consists of 5% soap, so the amount of soap from Passion Cook is $0.05x_2$
- Total soap required is at least 20 kg, so $0.10x_1 + 0.05x_2 \geq 20$
- Fruity Loop consists of 6% citric acid, so the amount of citric acid from Fruity Loop is $0.06x_1$
- Passion Cook consists of 10% citric acid, so the amount of citric acid from Passion Cook is $0.10x_2$
- Total citric acid required is at least 15 kg, so $0.06x_1 + 0.10x_2 \geq 15$
- Non-negativity constraints: $x_1 \geq 0, x_2 \geq 0$

## Gurobi Code

```python
import gurobi

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

    # Define variables
    x1 = model.addVar(name="Fruity_Loop", lb=0, ub=gurobi.GRB.INFINITY)
    x2 = model.addVar(name="Passion_Cook", lb=0, ub=gurobi.GRB.INFINITY)

    # Objective function: minimize total cost
    model.setObjective(6 * x1 + 5 * x2, gurobi.GRB.MINIMIZE)

    # Constraints
    model.addConstr(0.10 * x1 + 0.05 * x2 >= 20, name="soap_requirement")
    model.addConstr(0.06 * x1 + 0.10 * x2 >= 15, name="citric_acid_requirement")

    # Optimize
    model.optimize()

    # Print solution
    if model.status == gurobi.GRB.OPTIMAL:
        print(f"Optimal production levels: Fruity Loop = {x1.varValue} kg, Passion Cook = {x2.varValue} kg")
        print(f"Minimum cost: ${model.objVal:.2f}")
    else:
        print("No optimal solution found")

solve_problem()
```