2003 AMC 10B Problem 4

The whole bed is $11$ feet wide ($4 + 7$) and $6$ feet tall ($1 + 5$). Its five rectangular regions have dimensions

• top right: $7 \times 3$, area $21$;
• upper left: $4 \times 5$, area $20$;
• lower right: $5 \times 3$, area $15$;
• bottom strip: $6 \times 1$, area $6$;
• the leftover middle piece: $2 \times 2$, area $4$.

Check: $21 + 20 + 15 + 6 + 4 = 66 = 11 \cdot 6$, the area of the whole bed.

Each region gets one kind of flower, one plant per square foot. To make the total as small as possible, the cheapest flower should cover the most ground and the most expensive the least, so match prices $1, 1.5, 2, 2.5, 3$ with areas $21, 20, 15, 6, 4$ in that order:
$$1(21) + 1.5(20) + 2(15) + 2.5(6) + 3(4) = 21 + 30 + 30 + 15 + 12 = 108.$$

The answer is $\boxed{\textbf{(A)}\ 108}$.

Minimizing $\sum (\text{price}) \cdot (\text{area})$ over all pairings is the rearrangement principle: sort one list ascending and the other descending. If any two flowers were "out of order," swapping them would lower the cost, so the sorted pairing is optimal. The only real work is reading the unlabeled lengths correctly from the figure and checking they tile the whole bed.

• Misreading the dimensions of a region from the figure (the middle $2 \times 2$ square is easy to miss) or matching costs to areas in the wrong order.
• Pairing the cheapest flower with the smallest region, which gives the maximum cost $156$, choice (E).