2008 AMC 8 Problem 17

Below is the professionally curated solution for Problem 17 of the 2008 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 8 solutions, or check the answer key.

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Concepts:rectangleareaoptimization

Difficulty rating: 1180

17.

Ms. Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of 5050 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?

 76 \ 76

 120 \ 120

 128 \ 128

 132 \ 132

 136 \ 136

Solution:

If the side lengths are ll and ww, then 2l+2w=502l+2w=50, so l+w=25l+w=25.

The smallest possible area comes from side lengths 11 and 2424, giving area 2424.

The largest possible area comes from the closest integer pair with sum 2525, namely 1212 and 1313, giving area 156156.

The difference is 15624=132156-24=132.

Thus, D is the correct answer.

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