2020 AMC 12B Problems/Problem 22

Problem

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

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

Solution 1

We proceed by using AM-GM. We get $\frac{(2^t-3t) + 3t}{2}$ $\ge \sqrt{(2^t-3t)(3t)}$. Thus, squaring gives us that $4^{t-1} \ge (2^t-3t)(3t)$. Remembering what we want to find, we divide both sides of the inequality by the positive amount of $\frac{1}{3\cdot4^t}$. We get the maximal values as $\frac{1}{12}$, and we are done.

Solution 2

Set $u = t2^{-t}$. Then the expression in the problem can be written as $$R = - 3t^24^{-t} + t2^{-t}= - 3u^2 + u = - 3 (u - 1/6)^2 + \frac{1}{12} \le \frac{1}{12} .$$ 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 $\textbf{(C)}\ \frac{1}{12}$.

Solution 3 (Calculus Needed)

We want to maximize $f(t) = \frac{(2^t-3t)t}{4^t} = \frac{t\cdot 2^t-3t^2}{4^t}$. We can use the first derivative test. Use quotient rule to get the following: $$\frac{(2^t + t\cdot \ln{2} \cdot 2^t - 6t)4^t - (t\cdot 2^t - 3t^2)4^t \cdot 2\ln{2}}{(4^t)^2} = 0 \implies 2^t + t\cdot \ln{2} \cdot 2^t - 6t = (t\cdot 2^t - 3t^2) 2\ln{2}$$ $$\implies 2^t + t\cdot \ln{2}\cdot 2^t - 6t = 2t\ln{2} \cdot 2^t - 6t^2 \ln{2}$$ $$\implies 2^t(1-t\cdot \ln{2}) = 6t(1 - t\cdot \ln{2}) \implies 2^t = 6t$$Therefore, we plug this back into the original equation to get $\boxed{\textbf{(C)} \frac{1}{12}}$

~awesome1st

Solution 4

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 5

Let the maximum value of the function be $m$. Then we have $$\frac{(2^t-3t)t}{4^t} = m \implies m2^{2t} - t2^t + 3t^2 = 0.$$ Solving for $2^{t}$, we see $$2^{t} = \frac{t}{2m} \pm \frac{\sqrt{t^2 - 12mt^2}}{2m} = \frac{t}{2m} \pm \frac{t\sqrt{1 - 12m}}{2m}.$$ We see that $1 - 12m \geq 0 \implies m \leq \frac{1}{12}.$ Therefore, the answer is $\boxed{\textbf{(C)}\ \frac{1}{12}}$.

Solution 6

Let $z=2^t.$ Then,

$$\frac{\left(2^t-3t\right)t}{4^t}=\frac{\left(z-3t\right)t}{z^2} = \frac{-3t^2+zt}{z^2}.$$

Upon inspection, the numerator of this expression grows at a relatively faster rate than the denominator, when $t$ is close to $0$.

As the numerator is a quadratic in $t$ with a negative leading coefficient, its maximum value occurs at $t=\frac{-z}{2\cdot -3}=\frac{z}{6},$ or when $6t=2^t.$ Therefore,

$$\frac{\left(2^t-3t\right)t}{4^t}=\frac{\left(6t-3t\right)t}{(6t)^2}=\frac{3t^2}{36t^2}=\boxed{\textbf{(C)}\ \frac{1}{12}}.$$

-Benedict T (countmath1)

Video Solution1

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Video Solution

Problem starts at 2:10 in this video: https://www.youtube.com/watch?v=5HRSzpdJaX0&t=130s

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