Difference between revisions of "1983 AIME Problems/Problem 9"
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=== Solution 2 === | === Solution 2 === | ||
− | 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>. | + | 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>. |
− | This results in <math>\dfrac{(3x\sinx-2)^2}{x\sinx}+12</math>. | + | This results in <math>\dfrac{(3x \sinx-2)^2}{x \sinx}+12</math>. |
− | Thus, if <math>3x\sinx-2=0</math>, then the minimum is obviously 12. We show this possible with the same methods in Solution 1; thus the answer is <math>012</math>. | + | Thus, if <math>3x \sinx-2=0</math>, then the minimum is obviously 12. We show this possible with the same methods in Solution 1; thus the answer is <math>012</math>. |
=== Solution 3 === | === Solution 3 === |
Revision as of 20:02, 14 August 2013
Problem
Find the minimum value of for .
Solution
Solution 1
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 2
We can rewrite the numerator to be a perfect square by adding $-\dfrac{12x \sinx}{x \sinx}$ (Error compiling LaTeX. ). Thus, we must also add back .
This results in $\dfrac{(3x \sinx-2)^2}{x \sinx}+12$ (Error compiling LaTeX. ).
Thus, if $3x \sinx-2=0$ (Error compiling LaTeX. ), then the minimum is obviously 12. We show this possible with the same methods in Solution 1; thus the answer is .
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 |