Difference between revisions of "1987 AIME Problems/Problem 4"
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*<math>x - 60 < 0</math>. Then <math>y = \frac{5}{4}x-60</math>. | *<math>x - 60 < 0</math>. Then <math>y = \frac{5}{4}x-60</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>. | + | 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)) = \boxed{480}</math>. |
== See also == | == See also == | ||
Revision as of 17:54, 14 February 2014
Problem
Find the area of the region enclosed by the graph of 
Solution
Since 
is nonnegative, 
. Solving this gives us two equations: 
. Thus, 
. The maximum and minimum y value is when 
, which is when 
and 
. Since the graph is symmetric about the y-axis, we just need casework upon 
. 
, so we break up the condition 
:
. Then 
.
. Then 
.
. Then 
.
. Then 
.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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
