Difference between revisions of "2020 AMC 12B Problems/Problem 22"
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<math>\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}</math> | <math>\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}</math> | ||
− | == | + | ==Solution 1== |
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 - \frac{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 - \frac{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>. | ||
− | == | + | ==Solution 2 (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: | ||
+ | <cmath> | ||
+ | \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} | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | \implies 2^t + t\cdot \ln{2}\cdot 2^t - 6t = 2t\ln{2} \cdot 2^t - 6t^2 \ln{2} | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | \implies 2^t(1-t\cdot \ln{2}) = 6t(1 - t\cdot \ln{2}) \implies 2^t = 6t | ||
+ | </cmath>Therefore, we plug this back into the original equation to get <math>\boxed{\textbf{(C)} \frac{1}{12}}</math> | ||
+ | |||
+ | ==Solution 3== | ||
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>. | ||
− | ==Solution | + | ==Solution 4 (Bash)== |
Take the derivative of this function and let the derivative equals to 0, then this gives you <math>2^t=6t</math>. Substitute it into the original function you can get <math>\boxed{C}</math>. | Take the derivative of this function and let the derivative equals to 0, then this gives you <math>2^t=6t</math>. Substitute it into the original function you can get <math>\boxed{C}</math>. | ||
Revision as of 10:02, 8 February 2020
Contents
Problem 22
What is the maximum value of for real values of
Solution 1
Set . Then the expression in the problem can be written as It is easy to see that is attained for some value of between and , thus the maximal value of is .
Solution 2 (Calculus Needed)
We want to maximize . We can use the first derivative test. Use quotient rule to get the following: Therefore, we plug this back into the original equation to get
Solution 3
First, substitute so that
Notice that
When seen as a function, is a synthesis function that has as its inner function.
If we substitute , the given function becomes a quadratic function that has a maximum value of when .
Now we need to check if can have the value of in the range of real numbers.
In the range of (positive) real numbers, function is a continuous function whose value gets infinitely smaller as gets closer to 0 (as also diverges toward negative infinity in the same condition). When , , which is larger than .
Therefore, we can assume that equals to when is somewhere between 1 and 2 (at least), which means that the maximum value of is .
Solution 4 (Bash)
Take the derivative of this function and let the derivative equals to 0, then this gives you . Substitute it into the original function you can get .
See Also
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 |
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