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

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==Problem 17==
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==Problem==
 
Ms.Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of <math>50</math> 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?
 
Ms.Osborne asks each student in her class to draw a rectangle with integer side lengths and a perimeter of <math>50</math> 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?
  
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\textbf{(D)}\ 132\qquad
 
\textbf{(D)}\ 132\qquad
 
\textbf{(E)}\ 136</math>
 
\textbf{(E)}\ 136</math>
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==Solution==
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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>.
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==Video Solution==
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https://youtu.be/9bVwSsWa8IY Soo, DRMS, NM
  
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2008|num-b=16|num-a=18}}
 
{{AMC8 box|year=2008|num-b=16|num-a=18}}
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{{MAA Notice}}

Latest revision as of 00:26, 2 May 2022

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 $156-24=\boxed{\textbf{(D)}\ 132}$.


Video Solution

https://youtu.be/9bVwSsWa8IY Soo, DRMS, NM


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

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