Let's denote the number of paver stones Laila needs for each type as follows:
- Let \( x \) be the number of paver 2 stones needed.
- Therefore, the number of paver 1 stones needed will be \( 2x \) (twice as many as paver 2).
- The number of paver 3 stones needed will be \( \frac{x}{2} \) (half as many as paver 1).
Now, let's calculate the total cost for each type of paver.
-
Cost of Paver 1:
- Each paver 1 costs $1.49.
- Total cost for paver 1: \[ \text{Total Cost for Paver 1} = 2x \times 1.49 = 2.98x \]
-
Cost of Paver 2:
- Each paver 2 costs $2.50.
- Total cost for paver 2: \[ \text{Total Cost for Paver 2} = x \times 2.50 = 2.50x \]
-
Cost of Paver 3:
- Each paver 3 costs $0.75.
- Total cost for paver 3: \[ \text{Total Cost for Paver 3} = \frac{x}{2} \times 0.75 = 0.375x \]
Now we can summarize the total costs for each paver in terms of \( x \):
- Total cost for Paver 1: \( 2.98x \)
- Total cost for Paver 2: \( 2.50x \)
- Total cost for Paver 3: \( 0.375x \)
Now we can analyze which paver has the lowest cost:
- Paver 1: \( 2.98x \)
- Paver 2: \( 2.50x \)
- Paver 3: \( 0.375x \)
Since all costs depend on \( x \), we can compare them directly without needing a specific \( x \) value:
Clearly, \( 0.375x < 2.50x < 2.98x \).
Thus, the best buy is Paver 3.