## Step 1: Define the symbolic variables
Let's define the symbolic variables for the problem:
- $x_1$ represents the number of small coffee pods.
- $x_2$ represents the number of large coffee pods.

## Step 2: Formulate the objective function
The profit per small coffee pod is $3, and the profit per large coffee pod is $5. The objective function to maximize profit is:
\[ \text{Maximize:} \quad 3x_1 + 5x_2 \]

## 3: Define the constraints
1. There are only 2000 grams of coffee available. Each small coffee pod requires 15 grams, and each large coffee pod requires 20 grams:
\[ 15x_1 + 20x_2 \leq 2000 \]
2. At least 4 times the amount of small coffee pods are needed than large coffee pods:
\[ x_1 \geq 4x_2 \]
3. At least 10 large coffee pods need to be made:
\[ x_2 \geq 10 \]
4. Non-negativity constraints (number of pods cannot be negative):
\[ x_1 \geq 0, \quad x_2 \geq 0 \]

## 4: Symbolic Representation
The symbolic representation of the problem is:
```json
{
'sym_variables': [('x1', 'small coffee pods'), ('x2', 'large coffee pods')],
'objective_function': '3*x1 + 5*x2',
'constraints': [
    '15*x1 + 20*x2 <= 2000',
    'x1 >= 4*x2',
    'x2 >= 10',
    'x1 >= 0',
    'x2 >= 0'
]
}
```

## 5: Gurobi Code
Now, let's write the Gurobi code in Python to solve this problem:
```python
import gurobi

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

    # Define the variables
    x1 = model.addVar(name="small_coffee_pods", lb=0, vtype=gurobi.GRB.INTEGER)
    x2 = model.addVar(name="large_coffee_pods", lb=0, vtype=gurobi.GRB.INTEGER)

    # Objective function
    model.setObjective(3 * x1 + 5 * x2, gurobi.GRB.MAXIMIZE)

    # Constraints
    model.addConstr(15 * x1 + 20 * x2 <= 2000, name="coffee_limit")
    model.addConstr(x1 >= 4 * x2, name="small_to_large_ratio")
    model.addConstr(x2 >= 10, name="min_large_pods")

    # Optimize the model
    model.optimize()

    # Print the results
    if model.status == gurobi.GRB.OPTIMAL:
        print(f"Optimal solution: small coffee pods = {x1.varValue}, large coffee pods = {x2.varValue}")
    else:
        print("No optimal solution found")

solve_coffee_pod_problem()
```