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

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==Solution==
 
==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 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
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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
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==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}}

Latest revision as of 19:59, 27 October 2024

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 (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

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