2008 AMC 8 Problems/Problem 17

Problem

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?

$\textbf{(A)}\ 76\qquad \textbf{(B)}\ 120\qquad \textbf{(C)}\ 128\qquad \textbf{(D)}\ 132\qquad \textbf{(E)}\ 136$

Solution

A rectangle's area is maximized when its length and width are equivalent, or the two side lengths are closest together in this case with integer lengths. This occurs with the sides $12 \times 13 = 156$. Likewise, the area is smallest when the side lengths have the greatest difference, which is $1 \times 24 = 24$. The difference in area is $156-24=\boxed{\textbf{(D)}\ 132}$.

See Also

2008 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AJHSME/AMC 8 Problems and Solutions

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