To solve this problem, we first need to define the symbolic representation of the variables and the objective function, along with the constraints given in the problem description.

Let's denote:
- \(x_1\) as the amount invested in the avocado industry,
- \(x_2\) as the amount invested in the kale industry.

The objective is to maximize the return on investment. Given that the returns are 5% for the avocado industry and 8% for the kale industry, the total return can be represented by the objective function:
\[ \text{Maximize: } 0.05x_1 + 0.08x_2 \]

The constraints based on the problem description are:
1. The total amount invested is $300,000:
\[ x_1 + x_2 = 300,000 \]
2. At least 30% of the money should be invested in the avocado industry:
\[ x_1 \geq 0.3 \times 300,000 \]
\[ x_1 \geq 90,000 \]
3. At most 60% of the money should be invested in the kale industry:
\[ x_2 \leq 0.6 \times 300,000 \]
\[ x_2 \leq 180,000 \]

Additionally, since we are dealing with investments, both \(x_1\) and \(x_2\) must be non-negative.

In symbolic representation form as requested:
```json
{
    'sym_variables': [('x1', 'amount invested in avocado industry'), ('x2', 'amount invested in kale industry')],
    'objective_function': '0.05*x1 + 0.08*x2',
    'constraints': ['x1 + x2 = 300000', 'x1 >= 90000', 'x2 <= 180000', 'x1 >= 0', 'x2 >= 0']
}
```

Now, let's write the Gurobi code in Python to solve this linear programming problem:

```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(lb=0, name="avocado_investment")
x2 = m.addVar(lb=0, name="kale_investment")

# Set the objective function
m.setObjective(0.05*x1 + 0.08*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(x1 + x2 == 300000, "total_investment")
m.addConstr(x1 >= 90000, "min_avocado")
m.addConstr(x2 <= 180000, "max_kale")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Avocado Investment: {x1.x}")
    print(f"Kale Investment: {x2.x}")
    print(f"Total Return: {m.ObjVal}")
else:
    print("No optimal solution found")
```