To solve Jake's optimization problem, we need to formulate a linear programming model. The goal is to maximize profit by determining the optimal number of superhero and cartoon figurines to produce given the constraints on printer time.

Let:
- \(x\) be the number of superhero figurines produced,
- \(y\) be the number of cartoon figurines produced.

The objective function, which represents the total profit, can be formulated as:
\[ \text{Maximize:} \quad 10x + 8y \]

Given the constraints on printer time:
- Each superhero figurine requires 5 hours on Printer A and each cartoon figurine requires 3 hours on Printer A. The total time used on Printer A must not exceed 1000 hours, so we have:
\[ 5x + 3y \leq 1000 \]
- Each superhero figurine requires 4 hours on Printer B and each cartoon figurine requires 7 hours on Printer B. The total time used on Printer B must not exceed 1000 hours, so we have:
\[ 4x + 7y \leq 1000 \]

Additionally, \(x\) and \(y\) must be non-negative since they represent quantities of items.

Now, let's translate this problem into Gurobi code in Python:

```python
from gurobipy import *

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

# Define variables
superhero_figurines = m.addVar(vtype=GRB.CONTINUOUS, name="superhero_figurines", lb=0)
cartoon_figurines = m.addVar(vtype=GRB.CONTINUOUS, name="cartoon_figurines", lb=0)

# Objective function: Maximize profit
m.setObjective(10*superhero_figurines + 8*cartoon_figurines, GRB.MAXIMIZE)

# Constraints
m.addConstr(5*superhero_figurines + 3*cartoon_figurines <= 1000, name="Printer_A_Time")
m.addConstr(4*superhero_figurines + 7*cartoon_figurines <= 1000, name="Printer_B_Time")

# Optimize model
m.optimize()

# Print solution
if m.status == GRB.OPTIMAL:
    print(f"Optimal number of superhero figurines: {superhero_figurines.x:.2f}")
    print(f"Optimal number of cartoon figurines: {cartoon_figurines.x:.2f}")
    print(f"Maximum profit: ${m.objVal:.2f}")
else:
    print("No optimal solution found")
```