To solve this 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 given information.

Let's define:
- $x_1$ as the amount of shrimp (in kg) used in the mixture.
- $x_2$ as the amount of beef (in kg) used in the mixture.

The objective is to minimize the total cost of the mixture. Given that shrimp costs $36 per kg and beef costs $40 per kg, the objective function can be written as:
\[ \text{Minimize: } 36x_1 + 40x_2 \]

Now, let's consider the constraints:
1. The mixture needs to contain at least 20 units of protein.
2. The mixture needs to contain at least 25 units of fat.

Given that per kilogram, shrimp contains 2.5 units of protein and beef contains 4 units of protein, the first constraint can be written as:
\[ 2.5x_1 + 4x_2 \geq 20 \]

For the second constraint, since per kilogram, shrimp contains 3 units of fat and beef contains 2.5 units of fat, we have:
\[ 3x_1 + 2.5x_2 \geq 25 \]

Additionally, we should consider non-negativity constraints for $x_1$ and $x_2$, since the amount of shrimp or beef cannot be negative:
\[ x_1 \geq 0 \]
\[ x_2 \geq 0 \]

Thus, our symbolic representation is:
```json
{
    'sym_variables': [('x1', 'amount of shrimp in kg'), ('x2', 'amount of beef in kg')],
    'objective_function': '36*x1 + 40*x2',
    'constraints': ['2.5*x1 + 4*x2 >= 20', '3*x1 + 2.5*x2 >= 25', 'x1 >= 0', 'x2 >= 0']
}
```

To solve this problem using Gurobi in Python, we can write the following code:

```python
from gurobipy import *

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

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

# Set the objective function
m.setObjective(36*x1 + 40*x2, GRB.MINIMIZE)

# Add constraints
m.addConstr(2.5*x1 + 4*x2 >= 20, "protein_constraint")
m.addConstr(3*x1 + 2.5*x2 >= 25, "fat_constraint")

# Optimize model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Shrimp: {x1.x} kg")
    print(f"Beef: {x2.x} kg")
    print(f"Total cost: ${36*x1.x + 40*x2.x}")
else:
    print("No optimal solution found.")
```