To solve this problem, we first need to define the decision variables and the constraints based on the given information.

Let's denote:
- \(x_A\) as the amount of Mix A (in kg) to buy.
- \(x_B\) as the amount of Mix B (in kg) to buy.

The objective is to minimize the total cost, which can be represented as \(20x_A + 25x_B\).

We have two main constraints:
1. The final mixture must contain at least 10 kg of sugar. Given that Mix A contains 10% sugar and Mix B contains 15% sugar, this constraint can be written as \(0.10x_A + 0.15x_B \geq 10\).
2. The final mixture must contain at least 50 kg of flour. Since Mix A contains 60% flour and Mix B contains 50% flour, this constraint can be represented as \(0.60x_A + 0.50x_B \geq 50\).

Additionally, we know that \(x_A \geq 0\) and \(x_B \geq 0\), since the breakfast place cannot buy a negative amount of either mix.

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

```python
from gurobipy import *

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

# Define the decision variables
x_A = m.addVar(vtype=GRB.CONTINUOUS, name="Mix_A", lb=0)
x_B = m.addVar(vtype=GRB.CONTINUOUS, name="Mix_B", lb=0)

# Set the objective function: minimize cost
m.setObjective(20*x_A + 25*x_B, GRB.MINIMIZE)

# Add constraints
m.addConstr(0.10*x_A + 0.15*x_B >= 10, "Sugar_Constraint")
m.addConstr(0.60*x_A + 0.50*x_B >= 50, "Flour_Constraint")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Buy {x_A.x} kg of Mix A")
    print(f"Buy {x_B.x} kg of Mix B")
    print(f"Total cost: ${m.objVal}")
else:
    print("No optimal solution found")

```