2003 AMC 10B Problem 4

P1-10~3 minPrint

Quadrilaterals & Polygon Areas Absolute Value & Inequalities Set up the equation/formula and compute; no special trick needed Consider the largest/smallest element or boundary case

Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the figure. She plants one flower per square foot in each region. Asters cost $\$1$ each, begonias $\$1.50$ each, cannas $\$2$ each, dahlias $\$2.50$ each, and Easter lilies $\$3$ each. What is the least possible cost, in dollars, for her garden?

Show solution without answering

Key insight

Find the five areas from the figure (21, 20, 15, 6, 4). To spend the least, put the cheapest flower in the biggest region and the priciest in the smallest.

Solution

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 $6$ .
Work out each region's sides, using the labels and subtraction:

Region	Sides	Area
top right	$7 \times 3$	$21$
top left	$4 \times 5$	$20$
bottom right	$5 \times 3$	$15$
bottom strip	$6 \times 1$	$6$
small middle piece	$2 \times 2$	$4$

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Introduction to Geometry, Chapter 9: Polygons (§9.3)
Introduction to Algebra, Chapter 15: More Inequalities (§15.1, §15.3, §15.4)