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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==
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>. Rembering what we want to find(divide by <math>4^t</math>), we get the maximal values as <math>\frac{1}{12}</math>, and we are done.
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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>. Rembering 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.
  
 
==Solution 2==
 
==Solution 2==
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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>.
 
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==
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==Solution 6 (Simple)==
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.
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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> as the original expression, 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.
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~ math31415926535
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==Solution 7 (Simpler)==
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Distribute the terms and simplify <math>f(t)=\frac{(2^t-3t)t}{4^t}</math> to get <math>f(t)=\frac{2^tt}{4^t}-\frac{3t^2}{4^t}</math>. From there, we can set <math>u=\frac{t}{2^t}</math>, and substituting <math>u</math> into <math>f(t)</math> gets us <math>f(t)=u-3u^2</math>. Since the vertex of a quadratic is <math>\frac{-b}{2a}</math>, the <math>u</math> value of the vertex is <math>\frac{-1}{2(-3)}=\frac{1}{6}</math>. From there, we can obtain the maximum value of <math>f(t)</math>, which is <math>(\frac{1}{6})-3(\frac{1}{6})^2=\frac{1}{12}</math>, so the answer is <math>\boxed{C=\frac{1}{12}}</math>.
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~ Randomlygenerated
  
~ math31415926535 
 
 
==Video Solution==
 
==Video Solution==
 
Problem starts at 2:10 in this video: https://www.youtube.com/watch?v=5HRSzpdJaX0
 
Problem starts at 2:10 in this video: https://www.youtube.com/watch?v=5HRSzpdJaX0

Latest revision as of 23:24, 24 October 2021

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 $4^{t-1} \ge (2^t-3t)(3t)$. Rembering 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 (Simple)

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 $-(3y^{2}-y)$ as the original expression, 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

Solution 7 (Simpler)

Distribute the terms and simplify $f(t)=\frac{(2^t-3t)t}{4^t}$ to get $f(t)=\frac{2^tt}{4^t}-\frac{3t^2}{4^t}$. From there, we can set $u=\frac{t}{2^t}$, and substituting $u$ into $f(t)$ gets us $f(t)=u-3u^2$. Since the vertex of a quadratic is $\frac{-b}{2a}$, the $u$ value of the vertex is $\frac{-1}{2(-3)}=\frac{1}{6}$. From there, we can obtain the maximum value of $f(t)$, which is $(\frac{1}{6})-3(\frac{1}{6})^2=\frac{1}{12}$, so the answer is $\boxed{C=\frac{1}{12}}$.

~ Randomlygenerated

Video Solution

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

-MistyMathMusic

See Also

2020 AMC 12B (ProblemsAnswer KeyResources)
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

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