To solve the optimization problem described, 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 given information.

Let's denote:
- $x_1$ as the number of large coffees made.
- $x_2$ as the number of small coffees made.

The objective is to maximize profit. Given that the profit per large coffee is $5 and per small coffee is $3, the objective function can be written as:
\[ \text{Maximize:} \quad 5x_1 + 3x_2 \]

The constraints are based on the availability of resources (coffee beans and milk) and the requirements for each type of coffee. Specifically:
- A large coffee takes 12 units of coffee beans, so $x_1$ large coffees will take $12x_1$ units.
- A small coffee takes 8 units of coffee beans, so $x_2$ small coffees will take $8x_2$ units.
- The total available coffee beans are 1000 units, leading to the constraint: $12x_1 + 8x_2 \leq 1000$.

Similarly, for milk:
- A large coffee takes 20 units of milk, so $x_1$ large coffees will take $20x_1$ units.
- A small coffee takes 15 units of milk, so $x_2$ small coffees will take $15x_2$ units.
- The total available milk is 1500 units, leading to the constraint: $20x_1 + 15x_2 \leq 1500$.

Additionally, we have non-negativity constraints since the number of coffees cannot be negative:
\[ x_1 \geq 0, \quad x_2 \geq 0 \]

Thus, the symbolic representation of the problem is:
```json
{
    'sym_variables': [('x1', 'number of large coffees'), ('x2', 'number of small coffees')],
    'objective_function': '5*x1 + 3*x2',
    'constraints': ['12*x1 + 8*x2 <= 1000', '20*x1 + 15*x2 <= 1500', 'x1 >= 0', 'x2 >= 0']
}
```

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

# Create a model
m = Model("Coffee Shop Optimization")

# Define variables
x1 = m.addVar(lb=0, vtype=GRB.CONTINUOUS, name="large_coffees")
x2 = m.addVar(lb=0, vtype=GRB.CONTINUOUS, name="small_coffees")

# Set the objective function
m.setObjective(5*x1 + 3*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(12*x1 + 8*x2 <= 1000, "coffee_beans")
m.addConstr(20*x1 + 15*x2 <= 1500, "milk")

# Optimize the model
m.optimize()

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