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

(→‎Solution 3)
(Cleaned up the solutions)
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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>.
  
−
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]]:
+
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>.
  
−
Therefore, the minimum value is <math>\boxed{012}</math>. This is reached when plugging in <math>2/3</math> for <math>x\sin{x}</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).
+
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]]).
  
 
=== Solution 2 ===
 
=== Solution 2 ===
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<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>. 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>.
+
<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. 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>.
  
 
=== Solution 4 (also uses calculus) ===
 
=== 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{df}{dx} = \frac{6y(3y+2)-(3y+2)^2}{y^2}</math>. We find the minimum value by setting this to 0. 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 result. 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}{\frac{2}{3}}-12 = \boxed{012}</math>.
+
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>.
  
 
== See Also ==
 
== See Also ==

Revision as of 18:40, 15 February 2019

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 $\boxed{012}$. 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. 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 $\frac{(3(\frac{2}{3})+2)^2}{y}-12 = \frac{16}{\left(\frac{2}{3}\right)}-12 = \boxed{012}$.

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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