Cost Optimization: Flooring with Tiles

Floor dimensions: $9.5 \text{ m} \times 11.5 \text{ m}$. Maximum number of $1 \text{ m} \times 1 \text{ m}$ tiles used without cutting = $9 \times 11 = 99$ tiles. Remaining area requires $0.5 \text{ m} \times 0.5 \text{ m}$ tiles: Using 1x1 tiles for $9 \text{ m} \times 11 \text{ m}$ area = $99 \times 100 = 9900$. For remaining borders: $9 \text{ m} \times 0.5 \text{ m} = 4.5$ sq m (requires 18 tiles of 0.5x0.5 = $18 \times 30 = 540$), $11 \text{ m} \times 0.5 \text{ m} = 5.5$ sq m (requires 22 tiles of 0.5x0.5 = $22 \times 30 = 660$), corner $0.5 \times 0.5$ (1 tile = 30). Total cost = $9900 + 1030 = 10930$.