Difference between revisions of "1983 AIME Problems/Problem 9"
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Let <math>y = x\sin{x}</math> and rewrite the expression as <math>f(y) = 9y + \frac{4}{y}</math>. To minimize <math>f(y)</math>, take the derivative of <math>f(y)</math> and set it equal to zero. The derivative of <math>f(y)</math>, using the [[Power Rule]] is | Let <math>y = x\sin{x}</math> and rewrite the expression as <math>f(y) = 9y + \frac{4}{y}</math>. To minimize <math>f(y)</math>, take the derivative of <math>f(y)</math> and set it equal to zero. The derivative of <math>f(y)</math>, using the [[Power Rule]] is | ||
− | <math>f'(y)</math> = <math>9 - 4y^{-2} | + | <math>f'(y)</math> = <math>9 - 4y^{-2}</math> |
− | < | + | <math>f'(y)</math> is zero only when <math>y = \frac{2}{3}</math> or <math>y = -\frac{2}{3}</math>. However, since <math>x \sin{x}</math> is always positive in the given domain, <math>y = \frac{2}{3}</math>. Therefore, <math>x\sin{x}</math> = <math>\frac{2}{3}</math>, and the answer is <math>\frac{(9)(\exp{\frac{2}{3}}{2} + 4}{\frac{2}{3}} = \boxed{012}</math>. |
== See also == | == See also == |
Revision as of 19:25, 5 February 2010
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
=
is zero only when or . 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 |