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1983 AIME Problems/Problem 9: Difference between revisions

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== Solution 2 ==
== Solution 2 ==


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. The derivative of <math>f(y)</math>, using the [[Power Rule]] is
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.  
 
The derivative of <math>f(y)</math>, using the Power Rule, is


<math>f'(y)</math> = <math>9 - 4y^{-2}</math>
<math>f'(y)</math> = <math>9 - 4y^{-2}</math>


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


== See also ==
== See also ==

Revision as of 19:35, 5 February 2010

Problem

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

Solution

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

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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