To tackle the given optimization problem, we first need to translate the natural language description into a symbolic representation. This involves defining variables, an objective function, and constraints based on the problem's requirements.

Let's define:
- $x_1$ as the number of acres of potatoes to be planted,
- $x_2$ as the number of acres of squash to be planted.

The objective is to maximize profit. The profit from an acre of potatoes is $700, and from an acre of squash is $144. Therefore, the objective function can be represented algebraically as:
\[ 700x_1 + 144x_2 \]

Now, let's consider the constraints:
1. **Land Availability**: The total land available is 1000 acres. Thus,
\[ x_1 + x_2 \leq 1000 \]
2. **Tractor Time**: There are at most 1200 hours of tractor time available. Potatoes require 20 hours per acre, and squash requires 23 hours per acre. So,
\[ 20x_1 + 23x_2 \leq 1200 \]
3. **Capital Availability**: The farming group has $26400 of capital available. Each acre of potatoes requires $10 of capital, and each acre of squash requires $110 of capital. Thus,
\[ 10x_1 + 110x_2 \leq 26400 \]
4. **Non-Negativity Constraints**: Since the number of acres cannot be negative, we also have:
\[ x_1 \geq 0, x_2 \geq 0 \]

Given this symbolic representation, we can now encapsulate our problem description in JSON format as requested:

```json
{
    'sym_variables': [('x1', 'acres of potatoes'), ('x2', 'acres of squash')],
    'objective_function': '700*x1 + 144*x2',
    'constraints': ['x1 + x2 <= 1000', '20*x1 + 23*x2 <= 1200', '10*x1 + 110*x2 <= 26400', 'x1 >= 0', 'x2 >= 0']
}
```

Now, let's proceed to implement this problem in Gurobi using Python:

```python
from gurobipy import *

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

# Define the variables
x1 = m.addVar(name='acres_of_potatoes', lb=0)
x2 = m.addVar(name='acres_of_squash', lb=0)

# Set the objective function: Maximize profit
m.setObjective(700*x1 + 144*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(x1 + x2 <= 1000, name='land_availability')
m.addConstr(20*x1 + 23*x2 <= 1200, name='tractor_time')
m.addConstr(10*x1 + 110*x2 <= 26400, name='capital_availability')

# Optimize the model
m.optimize()

# Print out the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Acres of Potatoes: {x1.x}")
    print(f"Acres of Squash: {x2.x}")
    print(f"Total Profit: ${700*x1.x + 144*x2.x:.2f}")
else:
    print("No optimal solution found")
```