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

Latest revision as of 19:09, 13 October 2024

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 $\boxed{(C) \frac{1}{12}}$ , 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)

Solution 7 (fast)

Note that $\dfrac{(2^t-3t)t}{4^t}=\dfrac{t\cdot 2^t}{4^t}-\dfrac{3t^2}{4^t}=\dfrac{t}{2^t}-3\left(\dfrac{t}{2^t}\right)^2.$ Let $x=\frac{t}{2^t}.$ Then, the expression becomes $x-3x^2,$ which is minimized at $x=\frac{1}{6}$ , giving a value of $\boxed{\textbf{(C)}\ \frac{1}{12}}.$

Note: $\frac{t}{2^t}=\frac{1}{6}$ is possible as $\frac{t}{2^t}$ is continuous and that $\frac{0}{2^0}=0$ and $\frac{2}{2^2}=\frac{1}{2}$ so by the Intermediate Value Theorem (or just by intuition), there must be a $t$ between $0$ and $2$ that satisfies $\frac{t}{2^t}=\frac{1}{6}.$

~BS2012

Video Solution1

https://youtu.be/c2_b18Lv7c4

~Education, the Study of Everything

Video Solution

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

-MistyMathMusic

@@ Line 6: / Line 6: @@
 ==Solution 1==
-We proceed by using AM-GM. We get <math>\frac{(2^t-3t) + 3t}{2}</math> <math>\ge \sqrt{(2^t-3t)(3t)}</math>. Thus, squaring gives us that <math>4^{t-1} \ge (2^t-3t)(3t)</math>. Remembering what we want to find, we divide both sides of the inequality by the positive amount of <math>\frac{1}{3\cdot4^t}</math>. We get the maximal values as <math>\frac{1}{12}</math>, and we are done.
+We proceed by using AM-GM. We get <math>\frac{(2^t-3t) + 3t}{2}</math> <math>\ge \sqrt{(2^t-3t)(3t)}</math>. Thus, squaring gives us that <math>4^{t-1} \ge (2^t-3t)(3t)</math>. Remembering what we want to find, we divide both sides of the inequality by the positive amount of <math>\frac{1}{3\cdot4^t}</math>. We get the maximal values as <math>\boxed{(C)
+\frac{1}{12}}</math>, and we are done.
 ==Solution 2==

2020 AMC 12B (Problems • Answer Key • Resources)
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

Page

Toolbox

Search