Difference between revisions of "2008 AMC 8 Problems/Problem 17"
(→Solution) |
MRENTHUSIASM (talk | contribs) (Tag: Undo) |
||
| Line 9: | Line 9: | ||
==Solution== | ==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 | + | 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 <math>12 \times 13 = 156</math>. Likewise, the area is smallest when the side lengths have the greatest difference, which is <math>1 \times 24 = 24</math>. The difference in area is <math>156-24=\boxed{\textbf{(D)}\ 132}</math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=16|num-a=18}} | {{AMC8 box|year=2008|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:25, 8 August 2021
Problem
Ms.Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of 
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?
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 
. Likewise, the area is smallest when the side lengths have the greatest difference, which is 
. The difference in area is 
.
See Also
| 2008 AMC 8 (Problems • Answer Key • Resources) | ||
| 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. 
