Difference between revisions of "2020 AMC 12B Problems/Problem 22"
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==Solution1== | ==Solution1== | ||
− | Set <math> u = t2^{-t}</math>. Then the expression in the problem can be written as <cmath>R = t2^{-t} - 3t^24^{-t} = u - 3u^2 = \frac{1}{12}- 3 (\frac{1}{6} - u)^2 \le \frac{1}{12} .</cmath> It is easy to see that <math>u | + | Set <math> u = t2^{-t}</math>. Then the expression in the problem can be written as <cmath>R = t2^{-t} - 3t^24^{-t} = u - 3u^2 = \frac{1}{12}- 3 (\frac{1}{6} - u)^2 \le \frac{1}{12} .</cmath> It is easy to see that <math>u =\frac{1}{6}</math> is attained for some value of <math>t</math> between <math>t = 0</math> and <math>t = 1</math>, thus the maximal value of <math>R</math> is <math>\textbf{(C)}\ \frac{1}{12}</math>. |
==Solution2== | ==Solution2== |
Revision as of 01:16, 8 February 2020
Problem 22
What is the maximum value of for real values of
Solution1
Set . Then the expression in the problem can be written as
It is easy to see that
is attained for some value of
between
and
, thus the maximal value of
is
.
Solution2
First, substitute so that
Notice that
When seen as a function, is a synthesis function that has
as its inner function.
If we substitute , the given function becomes a quadratic function that has a maximum value of
when
.
Now we need to check that can have the value of
in the range of real numbers.
In the range of (positive) real numbers, function is a continuous function whose value gets infinitely smaller as
gets closer to 0 (as
also diverges toward negative infinity in the same condition). When
,
, which is larger than
.
Therefore, we can assume that equals to
when
is somewhere between 1 and 2 (at least), which means that the maximum value of
is
.