Problem 17 · 2008 AMC 8
Medium
Algebra & Patterns
fixed-perimeter-areamax-min
Ms. Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of 50 units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles?
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Answer: D — 132.
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Hint 1 of 2
Perimeter 50 fixes l + w = 25 — the two sides always add to 25, you only get to choose how to split it.
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Hint 2 of 2
For a fixed sum, the product (area) is biggest when the two sides are nearly equal and smallest when they're as lopsided as allowed.
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Approach: fixed sum ⇒ balanced is biggest, lopsided is smallest
- Half the perimeter is l + w = 25. With that sum fixed, area l·w is largest when the sides are as close as possible: 12 × 13 = 156.
- It's smallest when the sides are most lopsided. Sides must be positive integers, so the extreme is 1 × 24 = 24.
- Difference: 156 − 24 = 132.
- Why this transfers: for any fixed perimeter, "square-ish" rectangles enclose the most area — the same reason a square beats every other rectangle of equal perimeter.
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