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

(Solution2)
Line 11: Line 11:
 
==Solution2==
 
==Solution2==
  
First, substitute <math>2^t = x (log_2{x} = t)</math> so that  
+
First, substitute <math>2^t = x (\log_2{x} = t)</math> so that  
 
<cmath>
 
<cmath>
\frac{(2^t-3t)t}{4^t} = \frac{xlog_2{x}-3(log_2{x})^2}{x^2}
+
\frac{(2^t-3t)t}{4^t} = \frac{x\log_2{x}-3(\log_2{x})^2}{x^2}
 
</cmath>
 
</cmath>
  
 
Notice that  
 
Notice that  
 
<cmath>
 
<cmath>
\frac{xlog_2{x}-3(log_2{x})^2}{x^2} = \frac{log_2{x}}{x}-3(\frac{log_2{x}}{x})^2.
+
\frac{x\log_2{x}-3(\log_2{x})^2}{x^2} = \frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2.
 
</cmath>
 
</cmath>
  
When seen as a function, <math>\frac{log_2{x}}{x}-3(\frac{log_2{x}}{x})^2</math> is a synthesis function that has <math>\frac{log_2{x}}{x}</math> as its inner function.
+
When seen as a function, <math>\frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2</math> is a synthesis function that has <math>\frac{\log_2{x}}{x}</math> as its inner function.
  
If we substitute <math>\frac{log_2{x}}{x} = p</math>, the given function becomes a quadratic function that has a maximum value of <math>\frac{1}{12}</math> when <math>p = \frac{1}{6}</math>.
+
If we substitute <math>\frac{\log_2{x}}{x} = p</math>, the given function becomes a quadratic function that has a maximum value of <math>\frac{1}{12}</math> when <math>p = \frac{1}{6}</math>.
  
  
Now we need to check if <math>\frac{log_2{x}}{x}</math> can have the value of <math>\frac{1}{6}</math> in the range of real numbers.
+
Now we need to check if <math>\frac{\log_2{x}}{x}</math> can have the value of <math>\frac{1}{6}</math> in the range of real numbers.
  
In the range of (positive) real numbers, function <math>\frac{log_2{x}}{x}</math> is a continuous function whose value gets infinitely smaller as <math>x</math> gets closer to 0 (as <math>log_2{x}</math> also diverges toward negative infinity in the same condition). When <math>x = 2</math>, <math>\frac{log_2{x}}{x} = \frac{1}{2}</math>, which is larger than <math>\frac{1}{6}</math>.
+
In the range of (positive) real numbers, function <math>\frac{\log_2{x}}{x}</math> is a continuous function whose value gets infinitely smaller as <math>x</math> gets closer to 0 (as <math>log_2{x}</math> also diverges toward negative infinity in the same condition). When <math>x = 2</math>, <math>\frac{\log_2{x}}{x} = \frac{1}{2}</math>, which is larger than <math>\frac{1}{6}</math>.
  
Therefore, we can assume that <math>\frac{log_2{x}}{x}</math> equals to <math>\frac{1}{6}</math> when <math>x</math> is somewhere between 1 and 2 (at least), which means that the maximum value of <math>\frac{(2^t-3t)t}{4^t}</math> is <math>\textbf{(C)}\ \frac{1}{12}</math>.
+
Therefore, we can assume that <math>\frac{\log_2{x}}{x}</math> equals to <math>\frac{1}{6}</math> when <math>x</math> is somewhere between 1 and 2 (at least), which means that the maximum value of <math>\frac{(2^t-3t)t}{4^t}</math> is <math>\boxed{\textbf{(C)}\ \frac{1}{12}}</math>.
  
 
==Solution 3 (Bash)==
 
==Solution 3 (Bash)==

Revision as of 10:00, 8 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 \[R =  - 3t^24^{-t} + t2^{-t}= - 3u^2 + u = - 3 (u - \frac{1}{6})^2 + \frac{1}{12} \le \frac{1}{12} .\] It is easy to see that $u =\frac{1}{6}$ is attained for some value of $t$ between $t = 0$ and $t = 1$, thus the maximal value of $R$ is $\textbf{(C)}\ \frac{1}{12}$.

Solution2

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

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

When seen as a function, $\frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2$ 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 if $\frac{\log_2{x}}{x}$ 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 $\boxed{\textbf{(C)}\ \frac{1}{12}}$.

Solution 3 (Bash)

Take the derivative of this function and let the derivative equals to 0, then this gives you $2^t=6t$. Substitute it into the original function you can get $\boxed{C}$.

See Also

2020 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS