## Problem Description and Formulation

The problem is a linear programming optimization problem. The goal is to maximize the net revenue of a nut farmer who grows almonds and pecans on 80 acres of land. The farmer has limited resources: $10,000 for maintenance costs and 275 days of labor.

Let's define the decision variables:

* `almonds`: the number of acres of almonds to grow
* `pecans`: the number of acres of pecans to grow

The objective function to maximize is the net revenue:

* Net revenue per acre of almonds: $500
* Net revenue per acre of pecans: $600

The constraints are:

* Total land: 80 acres
* Maintenance costs: $10,000
* Labor: 275 days
* Non-negativity: `almonds >= 0` and `pecans >= 0`

## Mathematical Formulation

The mathematical formulation of the problem is:

Maximize:
`500 * almonds + 600 * pecans`

Subject to:
`almonds + pecans <= 80` (land constraint)
`200 * almonds + 250 * pecans <= 10000` (maintenance cost constraint)
`1.5 * almonds + 3 * pecans <= 275` (labor constraint)
`almonds >= 0` and `pecans >= 0` (non-negativity constraints)

## Gurobi Code

```python
import gurobi

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

# Define the decision variables
almonds = model.addVar(lb=0, name="almonds")
pecans = model.addVar(lb=0, name="pecans")

# Objective function: maximize net revenue
model.setObjective(500 * almonds + 600 * pecans, gurobi.GRB.MAXIMIZE)

# Land constraint: almonds + pecans <= 80
model.addConstr(almonds + pecans <= 80, name="land_constraint")

# Maintenance cost constraint: 200 * almonds + 250 * pecans <= 10000
model.addConstr(200 * almonds + 250 * pecans <= 10000, name="maintenance_cost_constraint")

# Labor constraint: 1.5 * almonds + 3 * pecans <= 275
model.addConstr(1.5 * almonds + 3 * pecans <= 275, name="labor_constraint")

# Optimize the model
model.optimize()

# Print the solution
if model.status == gurobi.GRB.OPTIMAL:
    print("Optimal solution:")
    print(f"Almonds: {almonds.varValue:.2f} acres")
    print(f"Pecans: {pecans.varValue:.2f} acres")
    print(f"Net revenue: ${model.objVal:.2f}")
else:
    print("No optimal solution found")
```