Difference between revisions of "1983 AIME Problems/Problem 9"
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
+ | We can rewrite the numerator to be a perfect square by adding <math>-\dfrac{12x\sinx}{x\sinx}</math>. Thus, we must also add back <math>12</math>. | ||
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+ | This results in <math>\dfrac{(3x\sinx-2)^2}{x\sinx}+12</math>. | ||
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+ | Thus, if 3x\sinx-2=0, then the minimum is obviously 12. We can show that this is possible. | ||
+ | |||
+ | Because <math>0<x<\pi</math>, <math>0<\sinx<1</math>. Thus, the value of <math>x\sinx = \frac{2}{3}</math> is obviously possible, thus the answer is <math>012</math>. | ||
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+ | === Solution 2 === | ||
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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Therefore, the minimum value is <math>\boxed{012}</math> (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> (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). | ||
− | === Solution | + | === Solution 3 === |
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. |
Revision as of 19:00, 14 August 2013
Problem
Find the minimum value of for .
Solution
Solution 1
We can rewrite the numerator to be a perfect square by adding $-\dfrac{12x\sinx}{x\sinx}$ (Error compiling LaTeX. ! Undefined control sequence.). Thus, we must also add back .
This results in $\dfrac{(3x\sinx-2)^2}{x\sinx}+12$ (Error compiling LaTeX. ! Undefined control sequence.).
Thus, if 3x\sinx-2=0, then the minimum is obviously 12. We can show that this is possible.
Because , $0<\sinx<1$ (Error compiling LaTeX. ! Undefined control sequence.). Thus, the value of $x\sinx = \frac{2}{3}$ (Error compiling LaTeX. ! Undefined control sequence.) is obviously possible, thus the answer is .
Solution 2
Let . We can rewrite the expression as .
Since and because , we have . So we can apply AM-GM:
The equality holds when .
Therefore, the minimum value is (when ; since is continuous and increasing on the interval and its range on that interval is from , by the Intermediate Value Theorem this value is attainable).
Solution 3
Let and rewrite the expression as , similar to the previous solution. To minimize , take the derivative of and set it equal to zero.
The derivative of , using the Power Rule, is
=
is zero only when or . It can further be verified that and are relative minima by finding the derivatives of other points near the critical points. However, since is always positive in the given domain, . Therefore, = , and the answer is .
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 |