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 $t?$

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

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 $4^{t-1} \ge (2^t-3t)(3t)$ . Rembering what we want to find(divide by $4^t$ ), 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

Set $x=2^t$ , now we get $\frac{(x-3t)t}{x^{2}}$ and we want to find the maximum value of that expression. And that expression can be further simplified to $(1-\frac{3t}{x})\frac{t}{x}$ , now lets set $\frac{t}{x}$ as $y$ and we get $(1-3y)y$ . Expand and we get $y-3y^{2}$ . We can make it simpler by rearranging and getting $-(3y^{2}-y)$ , so, now we need to find the minimum value of $3y^{2}-y$ and take the negative of that and get our answer. So, we complete the square on $3y^{2}-y$ to get it as $(y\sqrt{3}-\frac{\sqrt{3}}{6})^{2}-\frac{1}{12}$ . And by the trivial inequality, any real number squared is at least $0$ , so, to minimize this expression, we just set the square as $0$ and you subtract $\frac{1}{12}$ from $0$ to get $-\frac{1}{12}$ as the minimum value of the expression and you take the negative of that to get $\boxed{C=\frac{1}{12}}$ as the final answer.

~ math31415926535

Video Solution

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

-MistyMathMusic

@@ Line 54: / Line 54: @@
 Solving for <math>2^{t}</math>, we see <cmath>2^{t} = \frac{t}{2m} \pm \frac{\sqrt{t^2 - 12mt^2}}{2m} = \frac{t}{2m} \pm \frac{t\sqrt{1 - 12m}}{2m}.</cmath> We see that <math>1 - 12m \geq 0 \implies m \leq \frac{1}{12}.</math> Therefore, the answer is <math>\boxed{\textbf{(C)}\ \frac{1}{12}}</math>.
+==solution 6==
+Set <math>x=2^t</math>, now we get <math>\frac{(x-3t)t}{x^{2}}</math> and we want to find the maximum value of that expression. And that expression can be further simplified to <math>(1-\frac{3t}{x})\frac{t}{x}</math>, now lets set <math>\frac{t}{x}</math> as <math>y</math> and we get <math>(1-3y)y</math>. Expand and we get <math>y-3y^{2}</math>. We can make it simpler by rearranging and getting <math>-(3y^{2}-y)</math>, so, now we need to find the minimum value of <math>3y^{2}-y</math> and take the negative of that and get our answer. So, we complete the square on <math>3y^{2}-y</math> to get it as <math>(y\sqrt{3}-\frac{\sqrt{3}}{6})^{2}-\frac{1}{12}</math>. And by the trivial inequality, any real number squared is at least <math>0</math>, so, to minimize this expression, we just set the square as <math>0</math> and you subtract <math>\frac{1}{12}</math> from <math>0</math> to get <math>-\frac{1}{12}</math> as the minimum value of the expression and you take the negative of that to get <math>\boxed{C=\frac{1}{12}}</math> as the final answer.
+~ math31415926535
 ==Video Solution==
 Problem starts at 2:10 in this video: https://www.youtube.com/watch?v=5HRSzpdJaX0

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