To solve this optimization problem using Gurobi, we first need to understand and possibly simplify or rephrase the constraints given in the natural language description. The objective is to minimize the function \(1 \times \text{black beans} + 7 \times \text{steaks}\), subject to several constraints related to iron content, cost, fiber content, and a specific linear constraint involving black beans and steaks.

The constraints can be summarized as follows:
- Iron from black beans and steaks must be at least 18 mg and at most 71 mg.
- The total cost of black beans and steaks must be at least $22 and at most $45.
- Fiber from both sources must total at least 11 grams and no more than 28 grams.
- A specific constraint \(10 \times \text{black beans} - \text{steaks} \geq 0\) must be satisfied.
- Black beans can be fractional, but steaks must be integer.

Given these conditions, let's construct the Gurobi model:

```python
from gurobipy import *

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

# Define variables
black_beans = m.addVar(lb=0, ub=GRB.INFINITY, vtype=GRB.CONTINUOUS, name="black_beans")
steaks = m.addVar(lb=0, ub=GRB.INFINITY, vtype=GRB.INTEGER, name="steaks")

# Objective function: minimize 1*black_beans + 7*steaks
m.setObjective(1*black_beans + 7*steaks, GRB.MINIMIZE)

# Constraints
# Iron constraint: black beans contribute 1 mg each, steaks contribute 9 mg each
m.addConstr(1*black_beans + 9*steaks >= 18, name="iron_min")
m.addConstr(1*black_beans + 9*steaks <= 71, name="iron_max")

# Cost constraint: black beans cost $3 each, steaks cost $7 each
m.addConstr(3*black_beans + 7*steaks >= 22, name="cost_min")
m.addConstr(3*black_beans + 7*steaks <= 45, name="cost_max")

# Fiber constraint: black beans contribute 1 gram each, steaks contribute 3 grams each
m.addConstr(1*black_beans + 3*steaks >= 11, name="fiber_min")
m.addConstr(1*black_beans + 3*steaks <= 28, name="fiber_max")

# Specific linear constraint
m.addConstr(10*black_beans - steaks >= 0, name="specific_constraint")

# Optimize model
m.optimize()

# Print solution
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Black beans: {black_beans.x}")
    print(f"Steaks: {steaks.x}")
    print(f"Objective function value: {m.objVal}")
else:
    print("No optimal solution found")
```