Difference between revisions of "2020 AMC 12B Problems/Problem 22"
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==Solution 1== | ==Solution 1== | ||
+ | We proceed by using AM-GM. We get <math>\frac{(2^t-3t) + 3t}{2} </math>\ge \sqrt((2^t-3t)(3t))<math>. Thus, squaring gives us that </math>4^{t-1} \ge (2^t-3t)(3t)<math>. Rembering that what we want to find(divide by </math>4^t), we get the maximal values as <math>\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>. | ||
− | ==Solution | + | ==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 | ||
− | ==Solution | + | ==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 |
Revision as of 01:33, 2 April 2020
Contents
Problem 22
What is the maximum value of for real values of
Solution 1
We proceed by using AM-GM. We get \ge \sqrt((2^t-3t)(3t))4^{t-1} \ge (2^t-3t)(3t)4^t), we get the maximal values as , and we are done.
Solution 2
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 3 (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
~awesome1st
Solution 4
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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.