## Problem Description and Formulation

The problem is a classic example of a linear programming problem. The ice cream truck wants to maximize its revenue by selling ice cream cones and cups, given the constraints on the available scoops of ice cream and grams of toppings.

Let's define the decision variables:

* `x`: number of ice cream cones to sell
* `y`: number of ice cream cups to sell

The objective function is to maximize the revenue:

* Revenue per ice cream cone: $3
* Revenue per ice cream cup: $3.50
* Total revenue: `3x + 3.50y`

The constraints are:

* Each ice cream cone requires 3 scoops of ice cream, and each cup requires 4 scoops. The truck has 500 scoops available: `3x + 4y <= 500`
* Each ice cream cone requires 5 grams of toppings, and each cup requires 6 grams. The truck has 1000 grams available: `5x + 6y <= 1000`
* Non-negativity constraints: `x >= 0`, `y >= 0`

## Gurobi Code

```python
import gurobi

# Create a new model
m = gurobi.Model()

# Define the decision variables
x = m.addVar(name="ice_cream_cones", lb=0, ub=gurobi.GRB.INFINITY)
y = m.addVar(name="ice_cream_cups", lb=0, ub=gurobi.GRB.INFINITY)

# Define the objective function
m.setObjective(3 * x + 3.50 * y, gurobi.GRB.MAXIMIZE)

# Define the constraints
m.addConstr(3 * x + 4 * y <= 500, name="scoops_constraint")
m.addConstr(5 * x + 6 * y <= 1000, name="toppings_constraint")

# Optimize the model
m.optimize()

# Print the solution
if m.status == gurobi.GRB.OPTIMAL:
    print("Optimal solution found.")
    print(f"Number of ice cream cones to sell: {x.varValue}")
    print(f"Number of ice cream cups to sell: {y.varValue}")
    print(f"Maximum revenue: ${m.objVal:.2f}")
else:
    print("No optimal solution found.")
```

This code defines the decision variables, objective function, and constraints, and then solves the model using Gurobi. The solution is printed out, including the number of ice cream cones and cups to sell, and the maximum revenue. If no optimal solution is found, a message indicating this is printed.