Difference between revisions of "2020 AMC 12B Problems/Problem 22"
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− | ==Problem | + | ==Problem== |
What is the maximum value of <math>\frac{(2^t-3t)t}{4^t}</math> for real values of <math>t?</math> | What is the maximum value of <math>\frac{(2^t-3t)t}{4^t}</math> for real values of <math>t?</math> | ||
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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>\boxed{(C) |
+ | \frac{1}{12}}</math>, and we are done. | ||
==Solution 2== | ==Solution 2== | ||
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==Solution 5== | ==Solution 5== | ||
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}{2m} \pm \frac{\sqrt{t^2 - 12mt^2}}{ | + | 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== | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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, | ||
+ | |||
+ | <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) | ||
+ | |||
+ | ==Solution 7 (fast)== | ||
+ | Note that | ||
+ | <cmath>\dfrac{(2^t-3t)t}{4^t}=\dfrac{t\cdot 2^t}{4^t}-\dfrac{3t^2}{4^t}=\dfrac{t}{2^t}-3\left(\dfrac{t}{2^t}\right)^2.</cmath> | ||
+ | Let <math>x=\frac{t}{2^t}.</math> Then, the expression becomes <math>x-3x^2,</math> which is minimized at <math>x=\frac{1}{6}</math>, giving a value of <math>\boxed{\textbf{(C)}\ \frac{1}{12}}.</math> | ||
+ | |||
+ | |||
+ | |||
+ | Note: <math>\frac{t}{2^t}=\frac{1}{6}</math> is possible as <math>\frac{t}{2^t}</math> is continuous and that <math>\frac{0}{2^0}=0</math> and <math>\frac{2}{2^2}=\frac{1}{2}</math> so by the Intermediate Value Theorem (or just by intuition), there must be a <math>t</math> between <math>0</math> and <math>2</math> that satisfies <math>\frac{t}{2^t}=\frac{1}{6}.</math> | ||
+ | |||
+ | ~BS2012 | ||
+ | |||
+ | ==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 |
Latest revision as of 19:09, 13 October 2024
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)
Solution 7 (fast)
Note that Let Then, the expression becomes which is minimized at , giving a value of
Note: is possible as is continuous and that and so by the Intermediate Value Theorem (or just by intuition), there must be a between and that satisfies
~BS2012
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.