# 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 = | + | 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 = - 3 (u - \frac{1}{6})^2 + \frac{1}{12} \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 01:13, 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 .