Difference between revisions of "2020 AMC 12B Problems/Problem 22"
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==Solution2== | ==Solution2== | ||
− | 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 |
<cmath> | <cmath> | ||
− | \frac{(2^t-3t)t}{4^t} = \frac{ | + | \frac{(2^t-3t)t}{4^t} = \frac{x\log_2{x}-3(\log_2{x})^2}{x^2} |
</cmath> | </cmath> | ||
Notice that | Notice that | ||
<cmath> | <cmath> | ||
− | \frac{ | + | \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. |
</cmath> | </cmath> | ||
− | When seen as a function, <math>\frac{log_2{x}}{x}-3(\frac{log_2{x}}{x})^2</math> is a synthesis function that has <math>\frac{log_2{x}}{x}</math> as its inner function. | + | When seen as a function, <math>\frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2</math> is a synthesis function that has <math>\frac{\log_2{x}}{x}</math> as its inner function. |
− | If we substitute <math>\frac{log_2{x}}{x} = p</math>, the given function becomes a quadratic function that has a maximum value of <math>\frac{1}{12}</math> when <math>p = \frac{1}{6}</math>. | + | If we substitute <math>\frac{\log_2{x}}{x} = p</math>, the given function becomes a quadratic function that has a maximum value of <math>\frac{1}{12}</math> when <math>p = \frac{1}{6}</math>. |
− | Now we need to check if <math>\frac{log_2{x}}{x}</math> can have the value of <math>\frac{1}{6}</math> in the range of real numbers. | + | Now we need to check if <math>\frac{\log_2{x}}{x}</math> can have the value of <math>\frac{1}{6}</math> in the range of real numbers. |
− | In the range of (positive) real numbers, function <math>\frac{log_2{x}}{x}</math> is a continuous function whose value gets infinitely smaller as <math>x</math> gets closer to 0 (as <math>log_2{x}</math> also diverges toward negative infinity in the same condition). When <math>x = 2</math>, <math>\frac{log_2{x}}{x} = \frac{1}{2}</math>, which is larger than <math>\frac{1}{6}</math>. | + | In the range of (positive) real numbers, function <math>\frac{\log_2{x}}{x}</math> is a continuous function whose value gets infinitely smaller as <math>x</math> gets closer to 0 (as <math>log_2{x}</math> also diverges toward negative infinity in the same condition). When <math>x = 2</math>, <math>\frac{\log_2{x}}{x} = \frac{1}{2}</math>, which is larger than <math>\frac{1}{6}</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>\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 3 (Bash)== | ==Solution 3 (Bash)== |
Revision as of 11:00, 8 February 2020
Problem 22
What is the maximum value of for real values of
Solution1
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 .
Solution2
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 3 (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.