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

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Find the minimum value of <math>\frac{9x^2\sin^2 x + 4}{x\sin x}</math> for <math>0 < x < \pi</math>.
 
Find the minimum value of <math>\frac{9x^2\sin^2 x + 4}{x\sin x}</math> for <math>0 < x < \pi</math>.
  
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== Solution ==
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== Solution 1 ==
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=== Solution 1 ===
 
 
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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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]]:
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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>
 
<cmath>9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12</cmath>
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The equality holds when <math>9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23</math>.
 
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>. This is reached when plugging in <math>2/3</math> for <math>x\sinx</math> in the original equation (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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Therefore, the minimum value is <math>\boxed{012}</math>. This is reached when we have <math>x \sin{x} = \frac{2}{3}</math> in the original equation (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>, this value of <math>\frac{2}{3}</math> is attainable by the [[Intermediate Value Theorem]]).
  
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=== Solution 2 ===
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== Solution 2 ==
 
We can rewrite the numerator to be a perfect square by adding <math>-\dfrac{12x \sin x}{x \sin x}</math>. Thus, we must also add back <math>12</math>.
 
We can rewrite the numerator to be a perfect square by adding <math>-\dfrac{12x \sin x}{x \sin x}</math>. Thus, we must also add back <math>12</math>.
  
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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>.
 
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>.
  
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=== Solution 3 ===
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== Solution 3 (uses calculus) ==
  
 
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.  
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<math>f'(y)</math> = <math>9 - 4y^{-2}</math>
 
<math>f'(y)</math> = <math>9 - 4y^{-2}</math>
  
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<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}</math> 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}</math>.
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<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}</math> are relative minima by finding the derivatives at other points near the critical points, or by checking that the second derivative <math>f''(y)=8y^{-3}</math> is positive. 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>.
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 +
== Solution 4 (also uses calculus) ==
 +
 
 +
As above, let <math>y = x\sin{x}</math>. Add <math>\frac{12y}{y}</math> to the expression and subtract <math>12</math>, giving <math>f(x) = \frac{(3y+2)^2}{y} - 12</math>. Taking the [[derivative]] of <math>f(x)</math> using the [[Chain Rule]] and [[Quotient Rule]], we have <math>\frac{\text{d}f(x)}{\text{d}x} = \frac{6y(3y+2)-(3y+2)^2}{y^2}</math>. We find the minimum value by setting this to <math>0</math>. Simplifying, we have <math>6y(3y+2) = (3y+2)^2</math> and <math>y = \pm{\frac{2}{3}} = x\sin{x}</math>. Since both <math>x</math> and <math>\sin{x}</math> are positive on the given interval, we can ignore the negative root. Plugging <math>y = \frac{2}{3}</math> into our expression for <math>f(x)</math>, we have <math>\frac{(3(\frac{2}{3})+2)^2}{y}-12 = \frac{16}{\left(\frac{2}{3}\right)}-12 = \boxed{012}</math>.
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== Solution 5 ==
 +
 
 +
Set <math>\frac{9x^2\sin^2 x + 4}{x\sin x}</math> equal to <math>y</math>. Then multiply by <math>x\sin x</math> on both sides to get <math>9x^2\sin^2 x + 4 = y\cdot x\sin x</math>.  We then subtract <math>yx\sin x</math> from both sides to get <math>9x^2\sin^2 x + 4 - yx\sin x = 0</math>.  This looks like a quadratic so set <math>z= x\sin x</math> and use quadratic equation on <math>9z^2 - yz + 4 = 0</math> to see that <math>z = \frac{y\pm\sqrt{y^2-144}}{18}</math>. We know that <math>y</math> must be an integer and as small as it can be, so <math>y</math> = 12. We plug this back in to see that <math>x\sin x = \frac{2}{3}</math> which we can prove works using methods from solution 1. This makes the answer <math>\boxed{012}</math>
 +
 
 +
 
 +
-awesomediabrine
 +
 
 +
== Solution 6 ==
 +
Seeing that we need to minimize, we think inequalities, and seeing squares, we think [[RMS-AM-GM-HM]]. From this inequality, we know that <math>\sqrt{\frac{(3x\sin x)^2+2^2}{2}} \geq \sqrt{(3x\sin x)(2)}</math>, with equality holding when <math>3x\sin x=2</math>. From this inequality, we can see the following:
 +
\begin{align*}
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\sqrt{\frac{(3x\sin x)^2+2^2}{2}} \geq \sqrt{(3x\sin x)(2)} \\
 +
\frac{9x^2\sin^2x+4}{2} \geq 6x\sin x \\
 +
\frac{9x^2\sin^2x+4}{x\sin x} \geq 12
 +
\end{align*}
 +
We can prove that the equality condition is possible as in Solution <math>1</math>. Thus, our answer is <math>\boxed{012}</math>.
 +
 
 +
==Video Solution==
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 +
https://youtu.be/WQaL5cPDVVo
 +
 
 +
~Lucas
  
 
== See Also ==
 
== See Also ==

Latest revision as of 18:34, 20 July 2024

Problem

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

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 $\boxed{012}$. This is reached when we have $x \sin{x} = \frac{2}{3}$ in the original equation (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 $0 \le x\sin x \le \frac{\pi}{2}$, this value of $\frac{2}{3}$ is attainable by the Intermediate Value Theorem).

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 (uses calculus)

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 at other points near the critical points, or by checking that the second derivative $f''(y)=8y^{-3}$ is positive. 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 $\frac{(9)(\frac{2}{3})^2 + 4}{\frac{2}{3}} = \boxed{012}$.

Solution 4 (also uses calculus)

As above, let $y = x\sin{x}$. Add $\frac{12y}{y}$ to the expression and subtract $12$, giving $f(x) = \frac{(3y+2)^2}{y} - 12$. Taking the derivative of $f(x)$ using the Chain Rule and Quotient Rule, we have $\frac{\text{d}f(x)}{\text{d}x} = \frac{6y(3y+2)-(3y+2)^2}{y^2}$. We find the minimum value by setting this to $0$. Simplifying, we have $6y(3y+2) = (3y+2)^2$ and $y = \pm{\frac{2}{3}} = x\sin{x}$. Since both $x$ and $\sin{x}$ are positive on the given interval, we can ignore the negative root. Plugging $y = \frac{2}{3}$ into our expression for $f(x)$, we have $\frac{(3(\frac{2}{3})+2)^2}{y}-12 = \frac{16}{\left(\frac{2}{3}\right)}-12 = \boxed{012}$.

Solution 5

Set $\frac{9x^2\sin^2 x + 4}{x\sin x}$ equal to $y$. Then multiply by $x\sin x$ on both sides to get $9x^2\sin^2 x + 4 = y\cdot x\sin x$. We then subtract $yx\sin x$ from both sides to get $9x^2\sin^2 x + 4 - yx\sin x = 0$. This looks like a quadratic so set $z= x\sin x$ and use quadratic equation on $9z^2 - yz + 4 = 0$ to see that $z = \frac{y\pm\sqrt{y^2-144}}{18}$. We know that $y$ must be an integer and as small as it can be, so $y$ = 12. We plug this back in to see that $x\sin x = \frac{2}{3}$ which we can prove works using methods from solution 1. This makes the answer $\boxed{012}$


-awesomediabrine

Solution 6

Seeing that we need to minimize, we think inequalities, and seeing squares, we think RMS-AM-GM-HM. From this inequality, we know that $\sqrt{\frac{(3x\sin x)^2+2^2}{2}} \geq \sqrt{(3x\sin x)(2)}$, with equality holding when $3x\sin x=2$. From this inequality, we can see the following: \begin{align*} \sqrt{\frac{(3x\sin x)^2+2^2}{2}} \geq \sqrt{(3x\sin x)(2)} \\ \frac{9x^2\sin^2x+4}{2} \geq 6x\sin x \\ \frac{9x^2\sin^2x+4}{x\sin x} \geq 12 \end{align*} We can prove that the equality condition is possible as in Solution $1$. Thus, our answer is $\boxed{012}$.

Video Solution

https://youtu.be/WQaL5cPDVVo

~Lucas

See Also

1983 AIME (Problems • Answer Key • Resources)
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Problem 8 		Followed by
Problem 10
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All AIME Problems and Solutions
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