To solve the given optimization problem, we first need to define the decision variables and the objective function. Let's denote the number of cans of Ruby paint as \(R\) and the number of cans of Sapphire paint as \(S\).

The objective is to minimize the total cost, which can be represented as:
\[ \text{Minimize:} \quad 12R + 15S \]

Given the constraints:
- A can of Ruby paint contains 2 units of dye, 4 units of thinner, and 5 units of water.
- A can of Sapphire paint contains 4 units of dye, 6 units of thinner, and 2 units of water.
- The minimum requirements for the new mixture are 15 units of dye, 20 units of thinner, and 18 units of water.

We can represent these constraints as follows:
\[ 2R + 4S \geq 15 \] (Dye constraint)
\[ 4R + 6S \geq 20 \] (Thinner constraint)
\[ 5R + 2S \geq 18 \] (Water constraint)

Also, \(R\) and \(S\) must be non-negative since they represent the number of cans.

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

```python
from gurobipy import *

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

# Decision variables
R = m.addVar(vtype=GRB.CONTINUOUS, name="Ruby_Paint", lb=0)
S = m.addVar(vtype=GRB.CONTINUOUS, name="Sapphire_Paint", lb=0)

# Objective function: Minimize the total cost
m.setObjective(12*R + 15*S, GRB.MINIMIZE)

# Constraints
m.addConstr(2*R + 4*S >= 15, name="Dye_Constraint")
m.addConstr(4*R + 6*S >= 20, name="Thinner_Constraint")
m.addConstr(5*R + 2*S >= 18, name="Water_Constraint")

# Optimize the model
m.optimize()

# Print the results
if m.status == GRB.OPTIMAL:
    print("Optimal solution found:")
    print(f"Ruby Paint: {R.x} cans")
    print(f"Sapphire Paint: {S.x} cans")
    print(f"Total Cost: ${12*R.x + 15*S.x}")
else:
    print("No optimal solution found.")
```