## Step 1: Define the symbolic representation of the problem
Let's denote the hours of running plant Alpha as $x_1$ and the hours of running plant Beta as $x_2$. The objective is to minimize the cost, which is $40x_1 + 100x_2$. The constraints are:
- $3x_1 + 5x_2 \geq 100$ (cement demand)
- $2x_1 + 4x_2 \geq 80$ (stucco demand)
- $x_1 \geq 0$ and $x_2 \geq 0$ (non-negativity constraints)

## Step 2: Convert the problem into a symbolic representation
The symbolic variables are:
- $x_1$ for hours of running plant Alpha
- $x_2$ for hours of running plant Beta

The objective function is: $40x_1 + 100x_2$

The constraints are:
- $3x_1 + 5x_2 \geq 100$
- $2x_1 + 4x_2 \geq 80$
- $x_1 \geq 0$
- $x_2 \geq 0$

## 3: Express the problem in the required JSON format
```json
{
    'sym_variables': [('x1', 'hours of running plant Alpha'), ('x2', 'hours of running plant Beta')],
    'objective_function': '40*x1 + 100*x2',
    'constraints': [
        '3*x1 + 5*x2 >= 100',
        '2*x1 + 4*x2 >= 80',
        'x1 >= 0',
        'x2 >= 0'
    ]
}
```

## 4: Implement the problem in Gurobi code
To solve this linear programming problem using Gurobi in Python, we will use the following code:

```python
import gurobi

def solve_production_planning_problem():
    # Create a new model
    model = gurobi.Model()

    # Define variables
    x1 = model.addVar(name="Alpha_hours", lb=0)  # hours of running plant Alpha
    x2 = model.addVar(name="Beta_hours", lb=0)   # hours of running plant Beta

    # Objective function: minimize 40*x1 + 100*x2
    model.setObjective(40 * x1 + 100 * x2, gurobi.GRB.MINIMIZE)

    # Constraints
    model.addConstr(3 * x1 + 5 * x2 >= 100, name="cement_demand")  # cement demand
    model.addConstr(2 * x1 + 4 * x2 >= 80, name="stucco_demand")  # stucco demand

    # Solve the model
    model.optimize()

    # Print the results
    if model.status == gurobi.GRB.OPTIMAL:
        print("Optimal solution found.")
        print(f"Hours of running plant Alpha: {x1.varValue}")
        print(f"Hours of running plant Beta: {x2.varValue}")
        print(f"Total cost: {model.objVal}")
    else:
        print("No optimal solution found.")

solve_production_planning_problem()
```