To solve the given optimization problem, we first need to convert the natural language description into a symbolic representation. This involves defining variables for the quantities of Juice A and Juice B to be produced, formulating the objective function that represents the total profit, and listing the constraints based on the availability of raspberries, blueberries, and blackberries.

Let's denote:
- \(x_1\) as the number of units of Juice A to be made,
- \(x_2\) as the number of units of Juice B to be made.

The objective function aims to maximize profit. Given that the profit per unit of Juice A is $5 and per unit of Juice B is $7, the objective function can be represented as:
\[ \text{Maximize:} \quad 5x_1 + 7x_2 \]

The constraints are based on the availability of each type of berry:
- Raspberries: \(20x_1 + 15x_2 \leq 2000\)
- Blueberries: \(10x_1 + 15x_2 \leq 1500\)
- Blackberries: \(10x_1 + 5x_2 \leq 1400\)

Additionally, the quantities of Juice A and Juice B to be made cannot be negative:
- \(x_1 \geq 0\)
- \(x_2 \geq 0\)

Thus, the symbolic representation of the problem is:
```json
{
    'sym_variables': [('x1', 'Juice A'), ('x2', 'Juice B')],
    'objective_function': '5*x1 + 7*x2',
    'constraints': [
        '20*x1 + 15*x2 <= 2000',
        '10*x1 + 15*x2 <= 1500',
        '10*x1 + 5*x2 <= 1400',
        'x1 >= 0',
        'x2 >= 0'
    ]
}
```

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

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

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

# Set the objective function
m.setObjective(5*x1 + 7*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(20*x1 + 15*x2 <= 2000, name="Raspberries")
m.addConstr(10*x1 + 15*x2 <= 1500, name="Blueberries")
m.addConstr(10*x1 + 5*x2 <= 1400, name="Blackberries")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Juice A: {x1.x}")
    print(f"Juice B: {x2.x}")
    print(f"Total Profit: ${5*x1.x + 7*x2.x:.2f}")
else:
    print("No optimal solution found")
```