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2020 AMC 12B Problems/Problem 22

Revision as of 01:29, 8 February 2020 by Fanyuchen20020715 (talk | contribs) (→‎Solution2)

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 \[\frac{(2^t-3t)t}{4^t} = \frac{xlog_2{x}-3(log_2{x})^2}{x^2}\]

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 $\textbf{(C)}\ \frac{1}{12}$.

Solution 3

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 $\boxed{C}$.

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