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Difference between revisions of "1983 AIME Problems/Problem 9"

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== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===
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We can rewrite the numerator to be a perfect square by adding <math>-\dfrac{12x\sinx}{x\sinx}</math>. Thus, we must also add back <math>12</math>.
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This results in <math>\dfrac{(3x\sinx-2)^2}{x\sinx}+12</math>.
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Thus, if 3x\sinx-2=0, then the minimum is obviously 12. We can show that this is possible.
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Because <math>0<x<\pi</math>, <math>0<\sinx<1</math>. Thus, the value of <math>x\sinx = \frac{2}{3}</math> is obviously possible, thus the answer is <math>012</math>.
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=== Solution 2 ===
 
Let <math>y=x\sin{x}</math>. We can rewrite the expression as <math>\frac{9y^2+4}{y}=9y+\frac{4}{y}</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>.
  
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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).
 
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).
  
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=== Solution 2 ===
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=== Solution 3 ===
  
 
Let <math>y = x\sin{x}</math> and rewrite the expression as <math>f(y) = 9y + \frac{4}{y}</math>, similar to the previous solution. To minimize <math>f(y)</math>, take the [[derivative]] of <math>f(y)</math> and set it equal to zero.  
 
Let <math>y = x\sin{x}</math> and rewrite the expression as <math>f(y) = 9y + \frac{4}{y}</math>, similar to the previous solution. To minimize <math>f(y)</math>, take the [[derivative]] of <math>f(y)</math> and set it equal to zero.  

Revision as of 20:00, 14 August 2013

Problem

Find the minimum value of $\frac{9x^2\sin^2 x + 4}{x\sin x}$ for $0 < x < \pi$.

Solution

Solution 1

We can rewrite the numerator to be a perfect square by adding $-\dfrac{12x\sinx}{x\sinx}$ (Error compiling LaTeX. ). Thus, we must also add back $12$.

This results in $\dfrac{(3x\sinx-2)^2}{x\sinx}+12$ (Error compiling LaTeX. ).

Thus, if 3x\sinx-2=0, then the minimum is obviously 12. We can show that this is possible.

Because $0<x<\pi$, $0<\sinx<1$ (Error compiling LaTeX. ). Thus, the value of $x\sinx = \frac{2}{3}$ (Error compiling LaTeX. ) is obviously possible, thus the answer is $012$.

Solution 2

Let $y=x\sin{x}$. We can rewrite the expression as $\frac{9y^2+4}{y}=9y+\frac{4}{y}$.

Since $x>0$ and $\sin{x}>0$ because $0< x<\pi$, we have $y>0$. So we can apply AM-GM:

\[9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12\]

The equality holds when $9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23$.

Therefore, the minimum value is $\boxed{012}$ (when $x\sin{x}=\frac23$; since $x\sin x$ is continuous and increasing on the interval $0 \le x \le \frac{\pi}{2}$ and its range on that interval is from $0 \le x\sin x \le \frac{\pi}{2}$, by the Intermediate Value Theorem this value is attainable).

Solution 3

Let $y = x\sin{x}$ and rewrite the expression as $f(y) = 9y + \frac{4}{y}$, similar to the previous solution. To minimize $f(y)$, take the derivative of $f(y)$ and set it equal to zero.

The derivative of $f(y)$, using the Power Rule, is

$f'(y)$ = $9 - 4y^{-2}$

$f'(y)$ is zero only when $y = \frac{2}{3}$ or $y = -\frac{2}{3}$. It can further be verified that $\frac{2}{3}$ and $-\frac{2}{3}$ are relative minima by finding the derivatives of other points near the critical points. However, since $x \sin{x}$ is always positive in the given domain, $y = \frac{2}{3}$. Therefore, $x\sin{x}$ = $\frac{2}{3}$, and the answer is $\frac{(9)(\frac{2}{3})^2 + 4}{\frac{2}{3}} = \boxed{012}$.

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
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