To solve the given optimization problem, we first need to convert the natural language description into a symbolic representation. This involves defining variables, an objective function, and constraints based on the problem statement.

Let's define:
- $x_1$ as the amount of Fruity Loop detergent produced in kg.
- $x_2$ as the amount of Passion Cook detergent produced in kg.

The objective is to minimize the total cost. The cost of producing Fruity Loop is $6 per kg, and the cost of producing Passion Cook is $5 per kg. Therefore, the objective function can be represented as:
\[ \text{Minimize: } 6x_1 + 5x_2 \]

The constraints are based on the requirements for soap and citric acid:
- At least 20 kg of soap must be used.
- At least 15 kg of citric acid must be used.

Given that Fruity Loop consists of 10% soap and 6% citric acid, and Passion Cook consists of 5% soap and 10% citric acid, we can formulate the constraints as follows:
- Soap constraint: $0.10x_1 + 0.05x_2 \geq 20$
- Citric acid constraint: $0.06x_1 + 0.10x_2 \geq 15$

Additionally, since the amount of detergent produced cannot be negative, we have:
- Non-negativity constraints: $x_1 \geq 0$, $x_2 \geq 0$

Therefore, the symbolic representation of the problem is:
```json
{
    'sym_variables': [('x1', 'Fruity Loop'), ('x2', 'Passion Cook')],
    'objective_function': '6*x1 + 5*x2',
    'constraints': ['0.10*x1 + 0.05*x2 >= 20', '0.06*x1 + 0.10*x2 >= 15', 'x1 >= 0', 'x2 >= 0']
}
```

To solve this problem using Gurobi in Python, we can use the following code:
```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(name="Fruity_Loop", lb=0)
x2 = m.addVar(name="Passion_Cook", lb=0)

# Set the objective function
m.setObjective(6*x1 + 5*x2, GRB.MINIMIZE)

# Add constraints
m.addConstr(0.10*x1 + 0.05*x2 >= 20, name="Soap_Constraint")
m.addConstr(0.06*x1 + 0.10*x2 >= 15, name="Citric_Acid_Constraint")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Fruity Loop: {x1.x} kg")
    print(f"Passion Cook: {x2.x} kg")
    print(f"Total Cost: ${6*x1.x + 5*x2.x}")
else:
    print("No optimal solution found")
```