Difference between revisions of "1983 AIME Problems/Problem 9"
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== Solution == | == Solution == | ||
| − | Let <math>y=x\sin{x}</math>. | + | Let <math>y=x\sin{x}</math>. We can rewrite the expression as <math>\frac{9y^2+4}{y}=9y+\frac{4}{y}</math>. |
| − | + | Since <math>x>0</math> and <math>\sin{x}>0</math> because <math>0< x<\pi</math>, we have <math>y>0</math>. So we can apply [[AM-GM]]: | |
| − | + | <cmath>9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12</cmath> | |
| − | + | The equality holds when <math>9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23</math>. | |
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| − | The equality holds when <math>9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23</math> | ||
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| + | Therefore, the minimum value is <math>\boxed{012}</math> (when <math>x\sin{x}=\frac23</math>; since <math>x\sin x</math> is continuous and increasing on the interval <math>0 \le x \le \frac{\pi}{2}</math> and its range on that interval is from <math>0 \le x\sin x \le \frac{\pi}{2}</math>, by the [[Intermediate Value Theorem]] this value is attainable). | ||
== See also == | == See also == | ||
Revision as of 16:41, 31 December 2007
Problem
Find the minimum value of 
for 
.
Solution
Let 
. We can rewrite the expression as 
.
Since 
and 
because 
, we have 
. So we can apply AM-GM:
![\[9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12\]](http://latex.artofproblemsolving.com/6/b/2/6b2f5f491347817a8130ca9e339d65f43be1d26d.png)
The equality holds when 
.
Therefore, the minimum value is 
(when 
; since 
is continuous and increasing on the interval 
and its range on that interval is from 
, by the Intermediate Value Theorem this value is attainable).
See also
| 1983 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
