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 unknowns (number of regular hot-dogs and premium hot-dogs), formulating the objective function based on the profit, and listing all constraints as described.

### Symbolic Representation:

- **Variables:**
  - \(x_1\): Number of regular hot-dogs made per day.
  - \(x_2\): Number of premium hot-dogs made per day.

- **Objective Function:** 
  The objective is to maximize profit. Given that each regular hot-dog yields a profit of $3 and each premium hot-dog yields a profit of $5, the objective function can be represented as:
  \[ \text{Maximize:} \quad 3x_1 + 5x_2 \]

- **Constraints:**
  1. Demand constraint for regular hot-dogs: \(x_1 \leq 100\)
  2. Demand constraint for premium hot-dogs: \(x_2 \leq 250\)
  3. Total production capacity: \(x_1 + x_2 \leq 300\)
  4. Non-negativity constraints: \(x_1 \geq 0, x_2 \geq 0\)

### Problem Description in JSON Format:

```json
{
  'sym_variables': [('x1', 'Number of regular hot-dogs'), ('x2', 'Number of premium hot-dogs')],
  'objective_function': '3*x1 + 5*x2',
  'constraints': ['x1 <= 100', 'x2 <= 250', 'x1 + x2 <= 300', 'x1 >= 0', 'x2 >= 0']
}
```

### Gurobi Code:

To solve this linear programming problem, we will use the Gurobi Python interface. The code below formulates the model based on our symbolic representation and solves it.

```python
from gurobipy import *

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

# Add variables
x1 = m.addVar(vtype=GRB.CONTINUOUS, name="regular_hotdogs", lb=0)
x2 = m.addVar(vtype=GRB.CONTINUOUS, name="premium_hotdogs", lb=0)

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

# Add constraints
m.addConstr(x1 <= 100, "regular_demand")
m.addConstr(x2 <= 250, "premium_demand")
m.addConstr(x1 + x2 <= 300, "total_capacity")

# Optimize model
m.optimize()

# Print solution
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Regular hot-dogs: {x1.x}")
    print(f"Premium hot-dogs: {x2.x}")
    print(f"Maximum profit: {m.objVal}")
else:
    print("No optimal solution found")
```