To solve this optimization problem, we will use linear programming techniques. The goal is to minimize the total cost of purchasing vanilla and mocha flavored instant coffees while meeting the minimum requirements for caffeine and sugar intake.

Let's denote:
- \(V\) as the number of packages of vanilla flavor coffee purchased.
- \(M\) as the number of packages of mocha flavor coffee purchased.

The objective function to minimize is the total cost, which can be represented as:
\[2V + 3M\]

The constraints based on the problem description are:
1. Caffeine constraint: \(2V + 3M \geq 60\) (to consume at least 60 units of caffeine).
2. Sugar constraint: \(2V + 5M \geq 50\) (to consume at least 50 units of sugar).
3. Non-negativity constraints: \(V \geq 0, M \geq 0\), since we cannot purchase a negative number of packages.

Now, let's translate these into Gurobi code in Python:

```python
from gurobipy import *

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

# Define the variables
V = m.addVar(vtype=GRB.CONTINUOUS, name="Vanilla", lb=0)
M = m.addVar(vtype=GRB.CONTINUOUS, name="Mocha", lb=0)

# Set the objective function
m.setObjective(2*V + 3*M, GRB.MINIMIZE)

# Add constraints
m.addConstr(2*V + 3*M >= 60, "Caffeine_Constraint")
m.addConstr(2*V + 5*M >= 50, "Sugar_Constraint")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print(f"Optimal solution: {V.varName} = {V.x}, {M.varName} = {M.x}")
    print(f"Total cost: ${2*V.x + 3*M.x:.2f}")
else:
    print("No optimal solution found.")
```