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> <math>\ge \sqrt{(2^t-3t)(3t)}</math>. Thus, squaring gives us that <math>4^{t-1} \ge (2^t-3t)(3t)</math>. | + | 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>. Remembering 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 | + | ==Solution 6== |
− | |||
− | + | Let <math>z=2^t.</math> Then, | |
− | + | <cmath>\frac{\left(2^t-3t\right)t}{4^t}=\frac{\left(z-3t\right)t}{z^2} = \frac{-3t^2+zt}{z^2}.</cmath> | |
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− | ~ | + | Upon inspection, the numerator of this expression grows at a relatively faster rate than the denominator, when <math>t</math> is close to <math>0</math>. |
+ | |||
+ | As the numerator is a quadratic in <math>t</math> with a negative leading coefficient, its maximum value occurs at <math>t=\frac{-z}{2\cdot -3}=\frac{z}{6},</math> or when <math>6t=2^t.</math> Therefore, | ||
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+ | <cmath>\frac{\left(2^t-3t\right)t}{4^t}=\frac{\left(6t-3t\right)t}{(6t)^2}=\frac{3t^2}{36t^2}=\boxed{\textbf{(C)}\ \frac{1}{12}}.</cmath> | ||
+ | |||
+ | |||
+ | -Benedict T (countmath1) | ||
+ | |||
+ | ==Video Solution1== | ||
+ | https://youtu.be/c2_b18Lv7c4 | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==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&t=130s |
-MistyMathMusic | -MistyMathMusic |
Revision as of 14:26, 15 March 2023
Contents
Problem
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 . Remembering what we want to find, we divide both sides of the inequality by the positive amount of . 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 .
Solution 6
Let Then,
Upon inspection, the numerator of this expression grows at a relatively faster rate than the denominator, when is close to .
As the numerator is a quadratic in with a negative leading coefficient, its maximum value occurs at or when Therefore,
-Benedict T (countmath1)
Video Solution1
~Education, the Study of Everything
Video Solution
Problem starts at 2:10 in this video: https://www.youtube.com/watch?v=5HRSzpdJaX0&t=130s
-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.