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

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

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 $\boxed{C}$ .

Revision as of 01:32, 8 February 2020 (view source) Fanyuchen20020715 (talk \| contribs) (→‎Solution 3) ← Older edit		Revision as of 06:18, 8 February 2020 (view source) Bluefortress24 (talk \| contribs) (→‎Solution2) Newer edit →
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−	Now we need to check ~~that~~ <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>.

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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