Difference between revisions of "1987 AIME Problems/Problem 4"
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== Problem == | == Problem == | ||
| − | Find the area of the region enclosed by the graph of <math>\displaystyle |x-60|+|y|=|x/4|.</math> | + | Find the [[area]] of the region enclosed by the [[graph]] of <math>\displaystyle |x-60|+|y|=|x/4|.</math> |
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== Solution == | == Solution == | ||
| − | {{ | + | {{image}} |
| + | Since <math>|y|</math> is [[nonnegative]], <math>|\frac{x}{4}| \ge |x - 60|</math>. Solving this gives us two equations: <math>\frac{x}{4} \ge x - 60\ and \ -\frac{x}{4} \le x - 60</math>. Thus, <math>48 \le x \le 80</math>. The [[maximum]] and [[minimum]] y value is when <math>|x - 60| = 0</math>, which is when <math>x = 60</math> and <math>y = \pm 15</math>. The area of the region enclosed by the graph is that of the quadrilateral defined by the points <math>(48,0),\ (60,15),\ (80,0), \ (60,-15)</math>. Breaking it up into triangles and solving, we get <math>2 \cdot \frac{1}{2}(80 - 48)(15 - (-15)) = 480</math>. | ||
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== See also == | == See also == | ||
* [[1987 AIME Problems]] | * [[1987 AIME Problems]] | ||
{{AIME box|year=1987|num-b=3|num-a=5}} | {{AIME box|year=1987|num-b=3|num-a=5}} | ||
Revision as of 17:44, 15 February 2007
Problem
Find the area of the region enclosed by the graph of 
Solution
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Since 
is nonnegative, 
. Solving this gives us two equations: 
. Thus, 
. The maximum and minimum y value is when 
, which is when 
and 
. The area of the region enclosed by the graph is that of the quadrilateral defined by the points 
. Breaking it up into triangles and solving, we get 
.
See also
| 1987 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
