AMC 8 · 2008 · #17

Grade 4 geometry-2d

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

area-rectanglesperimetersystematic-enumerationoptimization-counting systematic-enumerationidentify-subproblems ↑ Prerequisites: area-rectanglesperimeter

📏 Short solution 💡 3 insights

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Problem

Each student draws a rectangle with whole-number side lengths and perimeter 50. They each compute the area. What is the largest possible area minus the smallest possible area?

Pick an answer.

(A)

(B)

120

(C)

128

(D)

132

(E)

136

AMC 8 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Extreme Principle

The perimeter pins down l + w = 25. The area l × w then depends on how we split 25 into two positive integers. Tool #14 (Extreme Principle) is the right tool because we want the max and min of a product whose two factors have a fixed sum: the product is biggest when the factors are as close as possible, and smallest when they are as far apart as possible. Tool #4 (Introduce a Variable) lets us name the sides l and w, and Tool #2 (Make a Systematic List) confirms the extremes by checking the endpoint pairs.

1STEP 1

Name the sides l and w; halving the perimeter turns 2(l+w)=50 into l+w=25, with both positive integers.

2(l + w) = 50 → l + w = 25, l,w ∈ {1,2,3,…}

2STEP 2

With l+w=25 fixed, area A=l(25-l) shrinks as the two factors spread apart — so it is biggest when they sit closest together.

A(l) = l × (25 - l); biggest when |l-w| is smallest, smallest when |l-w| is largest.

3STEP 3

The closest pair (12,13) gives the largest area 156; the farthest pair (1,24) gives the smallest area 24.

Pairs (l,w) with l ≤ w: (1,24),(2,23),…,(12,13). Largest area: 12 × 13 = 156. Smallest area: 1 × 24 = 24.

4STEP 4

Subtract the smallest area from the largest to get the difference the problem asks for.

156 - 24 = 132 → (D)

Answer

132

The answer 132 matches choice (D). Spot-check the trend with the middle of the list: (10,15) gives 150, (11,14) gives 154, (12,13) gives 156 — the product climbs as the two sides get closer, then there is no l=w case since 25 is odd. On the other end, (1,24)=24 and (2,23)=46 confirm the smallest area is at the most lopsided pair. So 156-24=132 is consistent.

💡Key takeaway

When two whole numbers add to a fixed total, their product is biggest when they are as close as possible and smallest when they are as far apart as possible.

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