Difference between revisions of "2008 AMC 8 Problems/Problem 17"

(Solution)
(Solution)
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 <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>.
+
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*13 = 156 Likewise, the area is smallest when the side lengths have the greatest difference, which is 1*24 = 24 The difference in area is 156-24=D
  
 
==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 17:49, 8 August 2021

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*13 = 156 Likewise, the area is smallest when the side lengths have the greatest difference, which is 1*24 = 24 The difference in area is 156-24=D

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