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

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==Problem 22==
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==Problem==
  
 
What is the maximum value of <math>\frac{(2^t-3t)t}{4^t}</math> for real values of <math>t?</math>
 
What is the maximum value of <math>\frac{(2^t-3t)t}{4^t}</math> for real values of <math>t?</math>
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==Solution 1==
 
==Solution 1==
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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.
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 +
==Solution 2==
  
 
Set <math> u = t2^{-t}</math>. Then the expression in the problem can be written as <cmath>R = - 3t^24^{-t} + t2^{-t}= - 3u^2 + u = - 3 (u - 1/6)^2 + \frac{1}{12} \le \frac{1}{12} .</cmath> It is easy to see that <math>u =\frac{1}{6}</math> is attained for some value of <math>t</math> between <math>t = 0</math> and <math>t = 1</math>, thus the maximal value of <math>R</math> is <math>\textbf{(C)}\ \frac{1}{12}</math>.
 
Set <math> u = t2^{-t}</math>. Then the expression in the problem can be written as <cmath>R = - 3t^24^{-t} + t2^{-t}= - 3u^2 + u = - 3 (u - 1/6)^2 + \frac{1}{12} \le \frac{1}{12} .</cmath> It is easy to see that <math>u =\frac{1}{6}</math> is attained for some value of <math>t</math> between <math>t = 0</math> and <math>t = 1</math>, thus the maximal value of <math>R</math> is <math>\textbf{(C)}\ \frac{1}{12}</math>.
  
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==Solution 2 (Calculus Needed)==
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==Solution 3 (Calculus Needed)==
  
 
We want to maximize <math>f(t) = \frac{(2^t-3t)t}{4^t} = \frac{t\cdot 2^t-3t^2}{4^t}</math>. We can use the first derivative test. Use quotient rule to get the following:
 
We want to maximize <math>f(t) = \frac{(2^t-3t)t}{4^t} = \frac{t\cdot 2^t-3t^2}{4^t}</math>. We can use the first derivative test. Use quotient rule to get the following:
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~awesome1st
 
~awesome1st
  
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==Solution 3==
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==Solution 4==
  
 
First, substitute <math>2^t = x (\log_2{x} = t)</math> so that  
 
First, substitute <math>2^t = x (\log_2{x} = t)</math> so that  
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Therefore, we can assume that <math>\frac{\log_2{x}}{x}</math> equals to <math>\frac{1}{6}</math> when <math>x</math> is somewhere between 1 and 2 (at least), which means that the maximum value of <math>\frac{(2^t-3t)t}{4^t}</math> is <math>\boxed{\textbf{(C)}\ \frac{1}{12}}</math>.
 
Therefore, we can assume that <math>\frac{\log_2{x}}{x}</math> equals to <math>\frac{1}{6}</math> when <math>x</math> is somewhere between 1 and 2 (at least), which means that the maximum value of <math>\frac{(2^t-3t)t}{4^t}</math> is <math>\boxed{\textbf{(C)}\ \frac{1}{12}}</math>.
  
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==Solution 4 (AM-GM)==
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==Solution 5==
 +
Let the maximum value of the function be <math>m</math>. Then we have <cmath>\frac{(2^t-3t)t}{4^t} = m \implies m2^{2t} - t2^t + 3t^2 = 0.</cmath>
 +
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>.
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 +
==Solution 6==
 +
 
 +
Let <math>z=2^t.</math> Then,
 +
 
 +
<cmath>\frac{\left(2^t-3t\right)t}{4^t}=\frac{\left(z-3t\right)t}{z^2} = \frac{-3t^2+zt}{z^2}.</cmath>
 +
 
 +
Upon inspection, the numerator of this expression grows at a relatively faster rate than the denominator, when <math>t</math> is close to <math>0</math>.
 +
 
 +
As the numerator is a quadratic in <math>t</math> with a negative leading coefficient, its maximum value occurs at <math>t=\frac{-z}{2\cdot -3}=\frac{z}{6},</math> or when <math>6t=2^t.</math> Therefore,
 +
 
 +
<cmath>\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}}.</cmath>
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 +
 
 +
-Benedict T (countmath1)
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==Solution 7 (fast)==
 +
Note that
 +
<cmath>\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.</cmath>
 +
Let <math>x=\frac{t}{2^t}.</math> Then, the expression becomes <math>x-3x^2,</math> which is minimized at <math>x=\frac{1}{6}</math>, giving a value of <math>\boxed{\textbf{(C)}\ \frac{1}{12}}.</math>
 +
 
 +
 
 +
 
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Note: <math>\frac{t}{2^t}=\frac{1}{6}</math> is possible as <math>\frac{t}{2^t}</math> is continuous and that <math>\frac{0}{2^0}=0</math> and <math>\frac{2}{2^2}=\frac{1}{2}</math> so by the Intermediate Value Theorem (or just by intuition), there must be a <math>t</math> between <math>0</math> and <math>2</math> that satisfies <math>\frac{t}{2^t}=\frac{1}{6}.</math>
 +
 
 +
~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
  
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Note that by AM-GM, <cmath>3t(2^t-3t) \leq \frac{2^{2t}}{4}</cmath>
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-MistyMathMusic
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So <cmath>\frac{t(2^t-3t)}{4^t} \leq \frac{1}{12}</cmath>
 
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Equality holds iff <math>2^t =6t</math> which happens for some <math>0<t<1</math> since <math>2^0>6 \times 0 </math> and <math>2^1<6 \times1</math>. So the answer is <math>\boxed{C. \frac{1}{12}}</math>
 
  
 
==See Also==
 
==See Also==

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

See Also

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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All AMC 12 Problems and Solutions

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