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

- \(L\) as the number of low-quality teas produced,
- \(M\) as the number of medium-quality teas produced,
- \(H\) as the number of high-quality teas produced.

The objective is to maximize profits. Given that the profit per low-quality tea is $1, per medium-quality tea is $3, and per high-quality tea is $5, the total profit \(P\) can be represented as:

\[ P = L + 3M + 5H \]

We have constraints based on the availability of rare additives and tea leaves. For rare additives, we have 200 units available, and for tea leaves, we have 400 units available. The consumption of these resources per type of tea is as follows:

- Low-quality tea: 2 units of rare additives, 6 units of tea leaves
- Medium-quality tea: 3 units of rare additives, 7 units of tea leaves
- High-quality tea: 4 units of rare additives, 8 units of tea leaves

Thus, the constraints can be formulated as:

1. Rare additives constraint: \(2L + 3M + 4H \leq 200\)
2. Tea leaves constraint: \(6L + 7M + 8H \leq 400\)

Additionally, we have non-negativity constraints since the production of teas cannot be negative:

- \(L \geq 0\)
- \(M \geq 0\)
- \(H \geq 0\)

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

```python
from gurobipy import *

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

# Define decision variables
L = m.addVar(name="Low_Quality_Tea", lb=0)
M = m.addVar(name="Medium_Quality_Tea", lb=0)
H = m.addVar(name="High_Quality_Tea", lb=0)

# Set the objective function
m.setObjective(L + 3*M + 5*H, GRB.MAXIMIZE)

# Add constraints
m.addConstr(2*L + 3*M + 4*H <= 200, name="Rare_Additives_Constraint")
m.addConstr(6*L + 7*M + 8*H <= 400, name="Tea_Leaves_Constraint")

# Optimize the model
m.optimize()

# Print results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found.")
    print(f"Low Quality Tea: {L.x}")
    print(f"Medium Quality Tea: {M.x}")
    print(f"High Quality Tea: {H.x}")
    print(f"Total Profit: {m.objVal}")
else:
    print("No optimal solution found.")
```