2020 AMC 12B Problems/Problem 22

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 $R = t2^{-t} - 3t^24^{-t} = u - 3u^2 = \frac{1}{12}- 3 (\frac{1}{6} - u)^2 \le \frac{1}{12} .$ It is easy to see that $u$ attains the value of $\frac{1}{6}$ for some value of $t$ between $0$ and $1$ , 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 $\textbf{(C)}\ \frac{1}{12}$ .

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2020 AMC 12B Problems/Problem 22

Problem 22

Solution1

Solution2