## Problem Description and Formulation

The kitchen company produces spoons, forks, and knives with rubber handles. The production requirements for each item are as follows:

- 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.
- Each knife requires 2 units of steel and 1 unit of rubber.

The company has limited resources:
- 400 units of steel available
- 500 units of rubber available

The revenue per item is:
- $2 per spoon
- $3 per fork
- $4 per knife

The goal is to maximize revenue by determining the optimal number of spoons, forks, and knives to produce.

## Mathematical Formulation

Let's denote the number of spoons, forks, and knives to be produced as \(S\), \(F\), and \(K\), respectively.

The objective function to maximize revenue is:
\[ \text{Maximize:} \quad 2S + 3F + 4K \]

Subject to the constraints:
\[ \text{Steel constraint:} \quad S + 1.5F + 2K \leq 400 \]
\[ \text{Rubber constraint:} \quad 2S + 1.5F + K \leq 500 \]
\[ \text{Non-negativity constraints:} \quad S \geq 0, F \geq 0, K \geq 0 \]

## Gurobi Code

```python
import gurobi

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

    # Define variables
    S = model.addVar(name="Spoons", lb=0, vtype=gurobi.GRB.CONTINUOUS)
    F = model.addVar(name="Forks", lb=0, vtype=gurobi.GRB.CONTINUOUS)
    K = model.addVar(name="Knives", lb=0, vtype=gurobi.GRB.CONTINUOUS)

    # Objective function: Maximize 2S + 3F + 4K
    model.setObjective(2 * S + 3 * F + 4 * K, gurobi.GRB.MAXIMIZE)

    # Steel constraint: S + 1.5F + 2K <= 400
    model.addConstr(S + 1.5 * F + 2 * K <= 400, name="Steel_Constraint")

    # Rubber constraint: 2S + 1.5F + K <= 500
    model.addConstr(2 * S + 1.5 * F + K <= 500, name="Rubber_Constraint")

    # Optimize
    model.optimize()

    # Print solution
    if model.status == gurobi.GRB.OPTIMAL:
        print("Optimal Solution:")
        print(f"Spoons: {S.varValue}")
        print(f"Forks: {F.varValue}")
        print(f"Knives: {K.varValue}")
        print(f"Max Revenue: {model.objVal}")
    else:
        print("No optimal solution found.")

maximize_revenue()
```