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</math> attains the value of <math>\frac{1}{6}</math> for some value of <math>t</math> between <math>0</math> and <math>1</math>, thus the maximal value of <math>R</math> is (C) | + | 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</math> attains the value of <math>\frac{1}{6}</math> for some value of <math>t</math> between <math>0</math> and <math>1</math>, thus the maximal value of <math>R</math> is <math>\textbf{(C)}\ \frac{1}{12}</math>. |
==Solution2== | ==Solution2== |
Revision as of 00:08, 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 attains the value of 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 .