To solve this problem, let's first break down the natural language description into a symbolic representation.

Let:
- $x_1$ be the number of pounds of sweet mix made.
- $x_2$ be the number of pounds of regular mix made.

The objective is to maximize profit. The profit per pound from selling the sweet mix is $3, and the profit per pound from selling the regular mix is $2. Thus, the objective function can be represented as:
\[ \text{Maximize:} \quad 3x_1 + 2x_2 \]

The constraints are based on the availability of butter popcorn and caramel popcorn:
- The sweet mix has 75% caramel popcorn and 25% butter popcorn.
- The regular mix has 50% caramel popcorn and 50% butter popcorn.
- There are 30 pounds of butter popcorn available.
- There are 40 pounds of caramel popcorn available.

Thus, the constraints can be represented as:
1. For butter popcorn: $0.25x_1 + 0.5x_2 \leq 30$
2. For caramel popcorn: $0.75x_1 + 0.5x_2 \leq 40$
3. Non-negativity constraints: $x_1 \geq 0, x_2 \geq 0$

In symbolic notation with natural language objects:
```json
{
    'sym_variables': [('x1', 'pounds of sweet mix made'), ('x2', 'pounds of regular mix made')],
    'objective_function': '3*x1 + 2*x2',
    'constraints': [
        '0.25*x1 + 0.5*x2 <= 30',
        '0.75*x1 + 0.5*x2 <= 40',
        'x1 >= 0',
        'x2 >= 0'
    ]
}
```

Now, let's write the Gurobi code in Python to solve this linear programming problem:

```python
from gurobipy import *

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

# Create variables
x1 = m.addVar(vtype=GRB.CONTINUOUS, name="sweet_mix")
x2 = m.addVar(vtype=GRB.CONTINUOUS, name="regular_mix")

# Set the objective function
m.setObjective(3*x1 + 2*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(0.25*x1 + 0.5*x2 <= 30, "butter_popcorn_constraint")
m.addConstr(0.75*x1 + 0.5*x2 <= 40, "caramel_popcorn_constraint")
m.addConstr(x1 >= 0, "non_negativity_sweet_mix")
m.addConstr(x2 >= 0, "non_negativity_regular_mix")

# Optimize model
m.optimize()

# Print solution
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Sweet mix: {x1.x} pounds")
    print(f"Regular mix: {x2.x} pounds")
    print(f"Maximum profit: ${3*x1.x + 2*x2.x}")
else:
    print("No optimal solution found.")
```