Difference between revisions of "2020 AMC 12B Problems/Problem 22"
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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> | 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}{m} \pm \frac{\sqrt{t^2 - 12mt^2}}{m} = \frac{t}{m} \pm \frac{t\sqrt{1 - 12m}}{m}.</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}{m} \pm \frac{\sqrt{t^2 - 12mt^2}}{m} = \frac{t}{m} \pm \frac{t\sqrt{1 - 12m}}{m}.</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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+ | ==Video Solution== | ||
+ | Problem starts at 2:10 in this video: https://www.youtube.com/watch?v=5HRSzpdJaX0 | ||
+ | |||
+ | -MistyMathMusic | ||
==See Also== | ==See Also== |
Revision as of 01:40, 6 August 2020
Contents
Problem 22
What is the maximum value of for real values of
Solution 1
We proceed by using AM-GM. We get . Thus, squaring gives us that . Rembering what we want to find(divide by ), 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 .
Solution 5
Let the maximum value of the function be . Then we have Solving for , we see We see that Therefore, the answer is .
Video Solution
Problem starts at 2:10 in this video: https://www.youtube.com/watch?v=5HRSzpdJaX0
-MistyMathMusic
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.