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

Let's define:
- $x_1$ as the number of packets of fruit gummy bears made,
- $x_2$ as the number of packets of sour gummy bears made.

The objective function, which is to maximize profit, can be represented as:
\[ \text{Maximize} \quad 1x_1 + 1.25x_2 \]

Given the constraints:
1. Time constraint: Each packet of fruit gummy bears takes 10 minutes, and each packet of sour gummy bears takes 15 minutes. The total available time is 2000 minutes.
\[ 10x_1 + 15x_2 \leq 2000 \]
2. Production limit for fruit gummy bears: At most 120 packets can be made.
\[ x_1 \leq 120 \]
3. Production limit for sour gummy bears: At most 70 packets can be made.
\[ x_2 \leq 70 \]
4. Non-negativity constraints, as the number of packets cannot be negative:
\[ x_1 \geq 0 \]
\[ x_2 \geq 0 \]

The symbolic representation in JSON format is:

```json
{
    'sym_variables': [('x1', 'number of packets of fruit gummy bears'), ('x2', 'number of packets of sour gummy bears')],
    'objective_function': 'Maximize 1*x1 + 1.25*x2',
    'constraints': ['10*x1 + 15*x2 <= 2000', 'x1 <= 120', 'x2 <= 70', 'x1 >= 0', 'x2 >= 0']
}
```

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

```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(name='fruit_gummy_bears', lb=0)  # Number of packets of fruit gummy bears
x2 = m.addVar(name='sour_gummy_bears', lb=0)   # Number of packets of sour gummy bears

# Set the objective function
m.setObjective(1*x1 + 1.25*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(10*x1 + 15*x2 <= 2000, name='time_constraint')
m.addConstr(x1 <= 120, name='fruit_production_limit')
m.addConstr(x2 <= 70, name='sour_production_limit')

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print(f"Optimal solution: {x1.varName} = {x1.x}, {x2.varName} = {x2.x}")
    print(f"Maximum profit: {m.objVal}")
else:
    print("No optimal solution found")
```