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 solar and finance calculators produced, expressing the objective function in terms of these variables, and formulating the constraints based on the available resources.

Let's denote:
- \(x_1\) as the number of solar calculators produced,
- \(x_2\) as the number of finance calculators produced.

The symbolic representation of the problem is given by:

```json
{
    'sym_variables': [('x1', 'number of solar calculators'), ('x2', 'number of finance calculators')],
    'objective_function': '12*x1 + 9*x2',
    'constraints': [
        '5*x1 + 3*x2 <= 150',  # Silicon constraint
        '4*x1 + 5*x2 <= 150',  # Plastic constraint
        '2*x1 + 3*x2 <= 70',   # Steel constraint
        'x1 >= 0',             # Non-negativity constraint for solar calculators
        'x2 >= 0'              # Non-negativity constraint for finance calculators
    ]
}
```

This representation captures the essence of the problem: maximizing profit (objective function) by producing solar and finance calculators (variables), subject to the limitations of silicon, plastic, and steel availability (constraints), and ensuring that the production quantities are non-negative.

Now, let's express this optimization problem in Gurobi code using Python:

```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(lb=0, vtype=GRB.CONTINUOUS, name="solar_calculators")
x2 = m.addVar(lb=0, vtype=GRB.CONTINUOUS, name="finance_calculators")

# Set the objective function
m.setObjective(12*x1 + 9*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(5*x1 + 3*x2 <= 150, "silicon_constraint")
m.addConstr(4*x1 + 5*x2 <= 150, "plastic_constraint")
m.addConstr(2*x1 + 3*x2 <= 70, "steel_constraint")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Production of solar calculators: {x1.x}")
    print(f"Production of finance calculators: {x2.x}")
    print(f"Maximum profit: {m.objVal}")
else:
    print("No optimal solution found")
```