To solve the given optimization problem, we first need to translate the natural language description into a symbolic representation. This involves defining variables for the quantities of interest (coffees and hot chocolates sold), formulating an objective function that represents the total profit, and listing all constraints based on the given limitations.

Let's denote:
- \(x_1\) as the number of coffees sold,
- \(x_2\) as the number of hot chocolates sold.

The objective function aims to maximize profit. Given that the profit on a coffee is $0.22 and on a hot chocolate is $0.14, we can write the objective function as:
\[ \text{Maximize: } 0.22x_1 + 0.14x_2 \]

Constraints are based on the minimum and maximum quantities of each item that can be sold, as well as the total number of items:
1. \( x_1 \geq 40 \) (at least 40 coffees must be sold),
2. \( x_2 \geq 20 \) (at least 20 hot chocolates must be sold),
3. \( x_1 \leq 60 \) (no more than 60 coffees can be made),
4. \( x_2 \leq 35 \) (no more than 35 hot chocolates can be made),
5. \( x_1 + x_2 \leq 75 \) (the total number of items cannot exceed 75).

Thus, the symbolic representation of the problem is:
```json
{
  'sym_variables': [('x1', 'number of coffees'), ('x2', 'number of hot chocolates')],
  'objective_function': '0.22*x1 + 0.14*x2',
  'constraints': ['x1 >= 40', 'x2 >= 20', 'x1 <= 60', 'x2 <= 35', 'x1 + x2 <= 75']
}
```

To solve this problem using Gurobi in Python, we can use the following code:

```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(lb=40, ub=60, vtype=GRB.INTEGER, name="coffees")
x2 = m.addVar(lb=20, ub=35, vtype=GRB.INTEGER, name="hot_chocolates")

# Set the objective function
m.setObjective(0.22*x1 + 0.14*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(x1 + x2 <= 75, "total_items_constraint")

# Optimize model
m.optimize()

# Print results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Coffees: {x1.x}")
    print(f"Hot Chocolates: {x2.x}")
    print(f"Total Profit: {m.objVal}")
else:
    print("No optimal solution found")
```