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 hoodies and sweaters to be made, formulating the objective function that represents the total profit, and listing the constraints based on the available resources (fabric and dye).

Let's define:
- $x_1$ as the number of hoodies to be made,
- $x_2$ as the number of sweaters to be made.

The objective function, which is the total profit, can be represented as:
\[20x_1 + 15x_2\]

The constraints are based on the availability of fabric and dye. For fabric, each hoodie requires 3 units, and each sweater requires 2 units, with a total of 500 units available. This gives us the constraint:
\[3x_1 + 2x_2 \leq 500\]

For dye, each hoodie requires 2 units, and each sweater requires 1.5 units, with a total of 300 units available. This gives us the constraint:
\[2x_1 + 1.5x_2 \leq 300\]

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

Now, let's represent this problem in the requested format:

```json
{
    'sym_variables': [('x1', 'number of hoodies'), ('x2', 'number of sweaters')], 
    'objective_function': '20*x1 + 15*x2', 
    'constraints': ['3*x1 + 2*x2 <= 500', '2*x1 + 1.5*x2 <= 300', 'x1 >= 0', 'x2 >= 0']
}
```

Next, we'll write the Gurobi code to solve this linear programming problem:

```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(vtype=GRB.CONTINUOUS, name="hoodies")
x2 = m.addVar(vtype=GRB.CONTINUOUS, name="sweaters")

# Set the objective function
m.setObjective(20*x1 + 15*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(3*x1 + 2*x2 <= 500, "fabric")
m.addConstr(2*x1 + 1.5*x2 <= 300, "dye")

# Optimize model
m.optimize()

# Print solution
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Number of hoodies: {x1.x}")
    print(f"Number of sweaters: {x2.x}")
    print(f"Total profit: {m.objVal}")
else:
    print("No optimal solution found")
```