To solve this optimization problem, we first need to define the decision variables and the objective function. Let's denote the number of spoons, forks, and knives produced as \(x_s\), \(x_f\), and \(x_k\) respectively. The objective is to maximize revenue, which can be calculated as \(2x_s + 3x_f + 4x_k\).

The constraints are based on the availability of steel and rubber:
1. Steel constraint: \(x_s + 1.5x_f + 2x_k \leq 400\)
2. Rubber constraint: \(2x_s + 1.5x_f + x_k \leq 500\)

All production quantities must be non-negative, so we also have:
- \(x_s \geq 0\)
- \(x_f \geq 0\)
- \(x_k \geq 0\)

Given these constraints and the objective function, we can formulate this problem as a linear programming problem.

Here is how you could implement this in Gurobi using Python:

```python
from gurobipy import *

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

# Define the decision variables
x_s = m.addVar(name="spoons", vtype=GRB.CONTINUOUS, lb=0)
x_f = m.addVar(name="forks", vtype=GRB.CONTINUOUS, lb=0)
x_k = m.addVar(name="knives", vtype=GRB.CONTINUOUS, lb=0)

# Define the objective function
m.setObjective(2*x_s + 3*x_f + 4*x_k, GRB.MAXIMIZE)

# Add constraints
m.addConstr(x_s + 1.5*x_f + 2*x_k <= 400, name="steel")
m.addConstr(2*x_s + 1.5*x_f + x_k <= 500, name="rubber")

# Optimize the model
m.optimize()

# Print solution
if m.status == GRB.OPTIMAL:
    print(f"Optimal solution: {x_s.varName} = {x_s.x}, {x_f.varName} = {x_f.x}, {x_k.varName} = {x_k.x}")
    print(f"Maximum revenue: {m.objVal}")
else:
    print("No optimal solution found")
```