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

Revision as of 23:11, 23 July 2006

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

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 $12$ (when $x\sin{x}=\frac23$ ).

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 == Problem ==
+Find the minimum value of <math>\frac{9x^2\sin^2 x + 4}{x\sin x}</math> for <math>0 < x < \pi</math>.
 == Solution ==
+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]]:
+<math>9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12</math>
+The equality holds when <math>9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23</math>
+Therefore, the minimum value is <math>12</math> (when <math>x\sin{x}=\frac23</math>).
+----
+* [[1983 AIME Problems/Problem 8|Previous Problem]]
+* [[1983 AIME Problems/Problem 10|Next Problem]]
+* [[1983 AIME Problems|Back to Exam]]
 == See also ==
-* [[1983 AIME Problems]]
+* [[AIME Problems and Solutions]]
+* [[American Invitational Mathematics Examination]]
+* [[Mathematics competition resources]]
+[[Category:Intermediate Algebra Problems]]
+[[Category:Intermediate Trigonometry Problems]]