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>t2^{-t} - | + | Set <math> u = t2^{-t}</math>. Then the expression in the problem can be written as <cmath>t2^{-t} - 3t^24^{-t} = u - 3u^2 = \frac{1}{12}- 3 (\frac{1}{6} - u)^2.</cmath> The maximal value is thus <math>\frac{1}{12}</math>. |
==Solution2== | ==Solution2== | ||
Revision as of 23:58, 7 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 ![\[t2^{-t} - 3t^24^{-t} = u - 3u^2 = \frac{1}{12}- 3 (\frac{1}{6} - u)^2.\]](http://latex.artofproblemsolving.com/8/d/5/8d570958e736db31cf00ff780a386ce783a8d87c.png)
The maximal value is thus 
.
Solution2
First, substitute 
so that
![\[\frac{(2^t-3t)t}{4^t} = \frac{xlog_2{x}-3(log_2{x})^2}{x^2}\]](http://latex.artofproblemsolving.com/9/5/4/9544bdce17f147049d2f78695837395163d625d5.png)
Notice that
![\[\frac{xlog_2{x}-3(log_2{x})^2}{x^2} = \frac{log_2{x}}{x}-3(\frac{log_2{x}}{x})^2.\]](http://latex.artofproblemsolving.com/0/f/d/0fd47fa5b14dfede694e99feeeed3bacc4f644f6.png)
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 
.
