To solve Jamie's diet problem, we first need to convert the natural language description into a symbolic representation of the optimization problem. Let's define the variables and the objective function, followed by the constraints.

### Symbolic Representation

- **Variables:**
  - \(x_1\): The amount of chicken (in units) that Jamie will eat.
  - \(x_2\): The amount of beef (in units) that Jamie will eat.

- **Objective Function:**
  The objective is to minimize the total cost. Given that chicken costs $3.4 per unit and beef costs $7.5 per unit, the objective function can be written as:
  \[
  \text{Minimize} \quad 3.4x_1 + 7.5x_2
  \]

- **Constraints:**
  1. **Protein Constraint:** Jamie needs at least 100 units of proteins. Since one unit of chicken has 10 units of proteins and one unit of beef has 30 units of proteins, the constraint can be written as:
     \[
     10x_1 + 30x_2 \geq 100
     \]
  2. **Fat Constraint:** Jamie needs at least 60 units of fat. Given that one unit of chicken has 6 units of fat and one unit of beef has 40 units of fat, the constraint can be written as:
     \[
     6x_1 + 40x_2 \geq 60
     \]
  3. **Non-Negativity Constraints:** Since Jamie cannot eat a negative amount of food, we have:
     \[
     x_1 \geq 0, \quad x_2 \geq 0
     \]

### JSON Representation

```json
{
  'sym_variables': [('x1', 'amount of chicken'), ('x2', 'amount of beef')],
  'objective_function': '3.4*x1 + 7.5*x2',
  'constraints': ['10*x1 + 30*x2 >= 100', '6*x1 + 40*x2 >= 60', 'x1 >= 0', 'x2 >= 0']
}
```

### Gurobi Code

```python
from gurobipy import *

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

# Add variables
x1 = m.addVar(vtype=GRB.CONTINUOUS, name="chicken", lb=0)
x2 = m.addVar(vtype=GRB.CONTINUOUS, name="beef", lb=0)

# Set the objective function
m.setObjective(3.4*x1 + 7.5*x2, GRB.MINIMIZE)

# Add constraints
m.addConstr(10*x1 + 30*x2 >= 100, "protein_constraint")
m.addConstr(6*x1 + 40*x2 >= 60, "fat_constraint")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Amount of chicken: {x1.x}")
    print(f"Amount of beef: {x2.x}")
    print(f"Total cost: {m.objVal}")
else:
    print("No optimal solution found")
```