To solve the given optimization problem, we first need to convert the natural language description into a symbolic representation. Let's denote the number of spoons, forks, and knives as $x_1$, $x_2$, and $x_3$ respectively.

The objective function is to maximize revenue. Given that the revenue per spoon is $2, the revenue per fork is $3, and the revenue per knife is $4, we can write the objective function as:
\[ \text{Maximize} \quad 2x_1 + 3x_2 + 4x_3 \]

The constraints are based on the availability of steel and rubber. Each spoon requires 1 unit of steel and 2 units of rubber, each fork requires 1.5 units of steel and 1.5 units of rubber, and each knife requires 2 units of steel and 1 unit of rubber. The company has available 400 units of steel and 500 units of rubber. Thus, the constraints can be written as:
\[ x_1 + 1.5x_2 + 2x_3 \leq 400 \] (steel constraint)
\[ 2x_1 + 1.5x_2 + x_3 \leq 500 \] (rubber constraint)
Also, the number of spoons, forks, and knives cannot be negative:
\[ x_1, x_2, x_3 \geq 0 \]

In symbolic notation with natural language objects, we have:
```json
{
    'sym_variables': [('x1', 'spoons'), ('x2', 'forks'), ('x3', 'knives')],
    'objective_function': '2*x1 + 3*x2 + 4*x3',
    'constraints': [
        'x1 + 1.5*x2 + 2*x3 <= 400',
        '2*x1 + 1.5*x2 + x3 <= 500',
        'x1 >= 0',
        'x2 >= 0',
        'x3 >= 0'
    ]
}
```

Now, let's implement this optimization problem using Gurobi in Python:
```python
from gurobipy import *

# Create a model
m = Model("Kitchen_Company")

# Define variables
x1 = m.addVar(vtype=GRB.CONTINUOUS, name="spoons")
x2 = m.addVar(vtype=GRB.CONTINUOUS, name="forks")
x3 = m.addVar(vtype=GRB.CONTINUOUS, name="knives")

# Set objective function
m.setObjective(2*x1 + 3*x2 + 4*x3, GRB.MAXIMIZE)

# Add constraints
m.addConstr(x1 + 1.5*x2 + 2*x3 <= 400, "steel_constraint")
m.addConstr(2*x1 + 1.5*x2 + x3 <= 500, "rubber_constraint")

# Optimize model
m.optimize()

# Print results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Spoons: {x1.x}")
    print(f"Forks: {x2.x}")
    print(f"Knives: {x3.x}")
else:
    print("No optimal solution found")
```