To solve the given optimization problem, 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 problem statement.

Let's define:
- $x_1$ as the number of mugs made,
- $x_2$ as the number of bowls made.

The objective is to maximize profit. Given that each mug brings a profit of $10 and each bowl brings a profit of $15, the objective function can be written as:
\[ \text{Maximize: } 10x_1 + 15x_2 \]

There are two main constraints:
1. Time constraint: The total time spent making mugs and bowls cannot exceed 1200 minutes. Since each mug takes 20 minutes to make and each bowl takes 30 minutes, we have:
\[ 20x_1 + 30x_2 \leq 1200 \]
2. Clay constraint: The artist only has enough clay to make a total of 50 items, which gives us:
\[ x_1 + x_2 \leq 50 \]

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

And because we can't make a fraction of an item, both $x_1$ and $x_2$ should be integers.

The symbolic representation of the problem is thus:
```json
{
  'sym_variables': [('x1', 'number of mugs'), ('x2', 'number of bowls')],
  'objective_function': '10*x1 + 15*x2',
  'constraints': ['20*x1 + 30*x2 <= 1200', 'x1 + x2 <= 50', 'x1 >= 0', 'x2 >= 0']
}
```

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

```python
from gurobipy import *

# Create a new model
model = Model("Pottery_Problem")

# Define variables
x1 = model.addVar(vtype=GRB.INTEGER, name="mugs")
x2 = model.addVar(vtype=GRB.INTEGER, name="bowls")

# Set the objective function
model.setObjective(10*x1 + 15*x2, GRB.MAXIMIZE)

# Add constraints
model.addConstr(20*x1 + 30*x2 <= 1200, "time_constraint")
model.addConstr(x1 + x2 <= 50, "clay_constraint")
model.addConstr(x1 >= 0, "non_neg_mugs")
model.addConstr(x2 >= 0, "non_neg_bowls")

# Optimize the model
model.optimize()

# Print the solution
if model.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Number of mugs: {x1.x}")
    print(f"Number of bowls: {x2.x}")
    print(f"Maximum profit: ${model.objVal}")
else:
    print("No optimal solution found")
```