To solve this optimization problem, we first need to define the variables and the objective function. Let's denote:

- $x_1$ as the number of Mixture 1 to sell
- $x_2$ as the number of Mixture 2 to sell

The objective is to maximize profit. Given that the profit per Mixture 1 is $10 and per Mixture 2 is $12, the objective function can be represented as:

Maximize: $10x_1 + 12x_2$

Now, let's consider the constraints based on the availability of almonds, pecans, and pistachios.

- For almonds: $20x_1 + 15x_2 \leq 1000$ (since Mixture 1 contains 20 grams of almonds and Mixture 2 contains 15 grams of almonds)
- For pecans: $30x_1 + 20x_2 \leq 1200$ (since Mixture 1 contains 30 grams of pecans and Mixture 2 contains 20 grams of pecans)
- For pistachios: $10x_1 + 25x_2 \leq 1100$ (since Mixture 1 contains 10 grams of pistachios and Mixture 2 contains 25 grams of pistachios)

Additionally, we have non-negativity constraints since the number of mixtures to sell cannot be negative:

- $x_1 \geq 0$
- $x_2 \geq 0$

So, in symbolic representation, we have:
```json
{
    'sym_variables': [('x1', 'Number of Mixture 1'), ('x2', 'Number of Mixture 2')],
    'objective_function': '10*x1 + 12*x2',
    'constraints': [
        '20*x1 + 15*x2 <= 1000',
        '30*x1 + 20*x2 <= 1200',
        '10*x1 + 25*x2 <= 1100',
        'x1 >= 0',
        'x2 >= 0'
    ]
}
```

Now, let's implement this problem using Gurobi in Python:
```python
from gurobipy import *

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

# Define variables
x1 = m.addVar(name="Mixture_1", vtype=GRB.CONTINUOUS, lb=0)
x2 = m.addVar(name="Mixture_2", vtype=GRB.CONTINUOUS, lb=0)

# Set objective function
m.setObjective(10*x1 + 12*x2, GRB.MAXIMIZE)

# Add constraints
m.addConstr(20*x1 + 15*x2 <= 1000, name="Almond_Constraint")
m.addConstr(30*x1 + 20*x2 <= 1200, name="Pecan_Constraint")
m.addConstr(10*x1 + 25*x2 <= 1100, name="Pistachio_Constraint")

# Optimize the model
m.optimize()

# Print results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Mixture 1: {x1.x}")
    print(f"Mixture 2: {x2.x}")
    print(f"Maximum Profit: {m.objVal}")
else:
    print("No optimal solution found")
```