1. **State the problem:** Mary has a parcel weighing 10.5 kg and needs to split it into two parcels, each weighing at most 6 kg, to ship to China. The shipping cost depends on the weight of each parcel, with given costs for 1 kg, 2 kg, and 4 kg. We need to find the least total cost for shipping both parcels. 2. **Analyze the constraints:** - Each parcel must weigh $\leq 6$ kg. - Total weight is 10.5 kg. - Costs given: - 1 kg costs $0.90$ - 2 kg costs $2.00$ - 4 kg costs $4.50$ 3. **Determine possible parcel weights:** Since the max per parcel is 6 kg, and total is 10.5 kg, possible splits are: - Parcel 1: $x$ kg, Parcel 2: $10.5 - x$ kg - Both $x$ and $10.5 - x$ must be $\leq 6$ 4. **Check feasible $x$ values:** - $x \leq 6$ - $10.5 - x \leq 6 \Rightarrow x \geq 4.5$ So $x$ is between 4.5 and 6 kg. 5. **Cost function:** We only have costs for 1, 2, and 4 kg. We assume costs scale linearly for other weights by combining these units. 6. **Calculate cost for parcel weights:** We try to split weights into sums of 1, 2, and 4 kg units to minimize cost. 7. **Try parcel 1 = 4.5 kg:** - 4 kg costs $4.50$ - 0.5 kg is not given, so we approximate by 1 kg cost $0.90$ (assuming partial kg costs same as 1 kg) - Total cost parcel 1: $4.50 + 0.90 = 5.40$ Parcel 2 = 6 kg: - 4 kg costs $4.50$ - 2 kg costs $2.00$ - Total cost parcel 2: $4.50 + 2.00 = 6.50$ Total cost: $5.40 + 6.50 = 11.90$ 8. **Try parcel 1 = 5 kg:** - 4 kg costs $4.50$ - 1 kg costs $0.90$ - Total parcel 1: $4.50 + 0.90 = 5.40$ Parcel 2 = 5.5 kg: - 4 kg costs $4.50$ - 1.5 kg approximated as 2 kg cost $2.00$ - Total parcel 2: $4.50 + 2.00 = 6.50$ Total cost: $5.40 + 6.50 = 11.90$ 9. **Try parcel 1 = 6 kg:** - 4 kg costs $4.50$ - 2 kg costs $2.00$ - Total parcel 1: $6.50$ Parcel 2 = 4.5 kg: - 4 kg costs $4.50$ - 0.5 kg approximated as 1 kg cost $0.90$ - Total parcel 2: $5.40$ Total cost: $6.50 + 5.40 = 11.90$ 10. **Conclusion:** All tested splits yield the same minimum total cost of $11.90$. **Final answer:** The least amount of money Mary has to pay is $11.90$.