Difference between revisions of "1983 AIME Problems/Problem 9"
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The equality holds when <math>9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23</math>. | The equality holds when <math>9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23</math>. | ||
− | Therefore, the minimum value is <math>\boxed{012}</math>. This is reached when plugging in <math>2/3</math> for <math>x\ | + | Therefore, the minimum value is <math>\boxed{012}</math>. This is reached when plugging in <math>2/3</math> for <math>x\sin{x}</math> in the original equation (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 2 === | === Solution 2 === |
Revision as of 12:53, 3 March 2017
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 . This is reached when plugging in
for
in the original equation (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 . Thus, we must also add back
.
This results in .
Thus, if , then the minimum is obviously
. 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 |