1983 AIME Problems/Problem 9
Contents
[hide]Problem
Find the minimum value of for .
Solution
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
Let and rewrite the expression as . To minimize , take the derivative of and set it equal to zero. The derivative of , using the Power Rule is
= f'(y)y = \frac{2}{3}y = -\frac{2}{3}x \sin{x}y = \frac{2}{3}x\sin{x}\frac{2}{3}\frac{(9)(\exp{\frac{2}{3}}{2} + 4}{\frac{2}{3}} = \boxed{012}$.
See also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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