1983 AIME Problems/Problem 9

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

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 $\frac{(9)(\frac{2}{3})^2 + 4}{\frac{2}{3}} = \boxed{012}$ .

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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1983 AIME Problems/Problem 9

Contents

Problem

Solution

Solution 1

Solution 2

See Also