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

(Solution 2)
(Solution 2)
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This results in <math>\dfrac{(3x \sin x-2)^2}{x \sin x}+12</math>.
 
This results in <math>\dfrac{(3x \sin x-2)^2}{x \sin x}+12</math>.
  
Thus, if <math>3x \sin x-2=0</math>, then the minimum is obviously 12. We show this possible with the same methods in Solution 1; thus the answer is <math>\boxed{012}</math>.
+
Thus, if <math>3x \sin x-2=0</math>, then the minimum is obviously <math>12</math>. We show this possible with the same methods in Solution 1; thus the answer is <math>\boxed{012}</math>.
  
 
=== Solution 3 ===
 
=== Solution 3 ===

Revision as of 13:04, 5 October 2013

Problem

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

Solution

Solution 1

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 2

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

This results in $\dfrac{(3x \sin x-2)^2}{x \sin x}+12$.

Thus, if $3x \sin x-2=0$, then the minimum is obviously $12$. We show this possible with the same methods in Solution 1; thus the answer is $\boxed{012}$.

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 (ProblemsAnswer KeyResources)
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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