2003 AMC 10B Problem 4

1. Read the outside of the figure. Across the top: $4 + 7 = 11$ feet. Down the left side: $1 + 5 = 6$ feet. So the bed is $11$ by $6$.
2. Work out each region's sides, using the labels and subtraction:

The middle piece is $6 - 4 = 2$ wide (the bottom strip is $6$ long, the top-left block is $4$ wide) and $5 - 3 = 2$ tall.
3. Check: $21 + 20 + 15 + 6 + 4 = 66$, and $11 \times 6 = 66$. Every square foot is counted once.
4. Each square foot gets one flower. To keep the bill small, the cheapest flower should fill the biggest region and the most expensive flower the smallest region. So pair prices $1, 1.50, 2, 2.50, 3$ with areas $21, 20, 15, 6, 4$:
$$1 \cdot 21 + 1.5 \cdot 20 + 2 \cdot 15 + 2.5 \cdot 6 + 3 \cdot 4 = 21 + 30 + 30 + 15 + 12 = 108.$$

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

If a pricier flower sat in a bigger region than a cheaper flower, swapping those two flowers would lower the total. So the cheapest plan has prices going up exactly as areas go down. Most of the real work here is reading the figure carefully; checking that the areas add up to the whole bed catches mistakes.

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 $\boxed{\textbf{(A)}\ 108}$.