Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AMC 12B Problems/Problem 22"

(→‎Solution1)
(→‎Solution1)
Line 7: Line 7:
 
==Solution1==
 
==Solution1==
  
−
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>.
+
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 (C) <math>\frac{1}{12}</math>.
  
 
==Solution2==
 
==Solution2==

Revision as of 23:59, 7 February 2020

Problem 22

What is the maximum value of $\frac{(2^t-3t)t}{4^t}$ for real values of $t?$

$\textbf{(A)}\ \frac{1}{16} \qquad\textbf{(B)}\ \frac{1}{15} \qquad\textbf{(C)}\ \frac{1}{12} \qquad\textbf{(D)}\ \frac{1}{10} \qquad\textbf{(E)}\ \frac{1}{9}$

Solution1

Set $u = t2^{-t}$. Then the expression in the problem can be written as \[t2^{-t} - 3t^24^{-t} = u - 3u^2 = \frac{1}{12}- 3 (\frac{1}{6} - u)^2.\] The maximal value is thus (C) $\frac{1}{12}$.

Solution2

First, substitute $2^t = x (log_2{x} = t)$ so that \[\frac{(2^t-3t)t}{4^t} = \frac{xlog_2{x}-3(log_2{x})^2}{x^2}\]

Notice that \[\frac{xlog_2{x}-3(log_2{x})^2}{x^2} = \frac{log_2{x}}{x}-3(\frac{log_2{x}}{x})^2.\]

When seen as a function, $\frac{log_2{x}}{x}-3(\frac{log_2{x}}{x})^2$ is a synthesis function that has $\frac{log_2{x}}{x}$ as its inner function.

If we substitute $\frac{log_2{x}}{x} = p$, the given function becomes a quadratic function that has a maximum value of $\frac{1}{12}$ when $p = \frac{1}{6}$.


Now we need to check that $\frac{log_2{x}}{x}$ can have the value of $\frac{1}{6}$ in the range of real numbers.

In the range of (positive) real numbers, function $\frac{log_2{x}}{x}$ is a continuous function whose value gets infinitely smaller as $x$ gets closer to 0 (as $log_2{x}$ also diverges toward negative infinity in the same condition). When $x = 2$, $\frac{log_2{x}}{x} = \frac{1}{2}$, which is larger than $\frac{1}{6}$.

Therefore, we can assume that $\frac{log_2{x}}{x}$ equals to $\frac{1}{6}$ when $x$ is somewhere between 1 and 2 (at least), which means that the maximum value of $\frac{(2^t-3t)t}{4^t}$ is $\textbf{(C)}\ \frac{1}{12}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_22&oldid=117331"