Competition · AMC preparation · step 4 of 4
AMC 8 · 2008 · #17
Grade 4 geometry-2dPick an answer.
AMC 8 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Turn the perimeter into a sum
Name the sides l and w; halving the perimeter turns 2(l+w)=50 into l+w=25, with both positive integers.
Using P = 2(l+w) for a rectangle is the Grade 3 perimeter standard. Dividing by 2 turns the constraint into a single sum.
3.MD.D.8Introduce A VariableUse the fixed-sum product rule
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.
Area of a rectangle is length × width (Grade 3). Among integer splits of 25, the closest pair is (12,13) and the farthest pair is (1,24).
With the two side lengths adding to the same fixed total, their product — the rectangle's area — is largest when the sides are as close together as possible and smallest when they are as far apart as possible.
▸ Why?
Line each pair up against the middle of the total: since the two sides add to 25, write them as 12.5 - g and 12.5 + g, where g is the gap out to each side; sides that are farther apart simply have a larger gap g.
▸ Why?
Multiplying those two expressions out collapses to the middle squared minus the gap squared: (12.5 - g)(12.5 + g) = 12.5² - g², so the area is one fixed number, 12.5², with g² taken away.
▸ Why?
That fixed amount 12.5² is split with no overlap into the area and the piece g²; because the whole stays the same size, a larger gap makes the g² piece take more and leaves the area smaller, while the smallest possible gap leaves the area with the most.
Compute the extreme cases
The closest pair (12,13) gives the largest area 156; the farthest pair (1,24) gives the smallest area 24.
Listing the pairs makes the pattern visible: products grow 24, 46, 66, … as the gap shrinks, peaking at 156 for (12,13).
3.MD.C.7Make A Systematic ListSubtract the two areas
Subtract the smallest area from the largest to get the difference the problem asks for.
The multi-step word problem ends in one subtraction (Grade 4): max area minus min area.
4.OA.A.3Extreme PrincipleWhen 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.
- Turn the perimeter into a sum
- Use the fixed-sum product rule
- Compute the extreme cases
- Subtract the two areas
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