2008 AMC 8 Problems/Problem 17

Revision as of 17:49, 29 August 2013 by Schen001 (talk | contribs)

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 it is shaped like a square, 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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png