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>. Rembering what we want to find | + | 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>. Rembering 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 (Simple)== |
− | Set <math>x=2^t</math>, now we get <math>\frac{(x-3t)t}{x^{2}}</math> and we want to find the maximum value of that expression. And that expression can be further simplified to <math>(1-\frac{3t}{x})\frac{t}{x}</math>, now lets set <math>\frac{t}{x}</math> as <math>y</math> and we get <math>(1-3y)y</math>. Expand and we get <math>y-3y^{2}</math>. We can make it simpler by rearranging and getting <math>-(3y^{2}-y)</math>, so, now we need to find the minimum value of <math>3y^{2}-y</math> and take the negative of that and get our answer. So, we complete the square on <math>3y^{2}-y</math> to get it as <math>(y\sqrt{3}-\frac{\sqrt{3}}{6})^{2}-\frac{1}{12}</math>. And by the trivial inequality, any real number squared is at least <math>0</math>, so, to minimize this expression, we just set the square as <math>0</math> and you subtract <math>\frac{1}{12}</math> from <math>0</math> to get <math>-\frac{1}{12}</math> as the minimum value of the expression and you take the negative of that to get <math>\boxed{C=\frac{1}{12}}</math> as the final answer. | + | Set <math>x=2^t</math>, now we get <math>\frac{(x-3t)t}{x^{2}}</math> and we want to find the maximum value of that expression. And that expression can be further simplified to <math>(1-\frac{3t}{x})\frac{t}{x}</math>, now lets set <math>\frac{t}{x}</math> as <math>y</math> and we get <math>(1-3y)y</math>. Expand and we get <math>y-3y^{2}</math>. We can make it simpler by rearranging and getting <math>-(3y^{2}-y)</math> as the original expression, so, now we need to find the minimum value of <math>3y^{2}-y</math> and take the negative of that and get our answer. So, we complete the square on <math>3y^{2}-y</math> to get it as <math>(y\sqrt{3}-\frac{\sqrt{3}}{6})^{2}-\frac{1}{12}</math>. And by the trivial inequality, any real number squared is at least <math>0</math>, so, to minimize this expression, we just set the square as <math>0</math> and you subtract <math>\frac{1}{12}</math> from <math>0</math> to get <math>-\frac{1}{12}</math> as the minimum value of the expression and you take the negative of that to get <math>\boxed{C=\frac{1}{12}}</math> as the final answer. |
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+ | ~ math31415926535 | ||
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+ | ==Solution 7 (Simpler)== | ||
+ | Distribute the terms and simplify <math>f(t)=\frac{(2^t-3t)t}{4^t}</math> to get <math>f(t)=\frac{2^tt}{4^t}-\frac{3t^2}{4^t}</math>. From there, we can set <math>u=\frac{t}{2^t}</math>, and substituting <math>u</math> into <math>f(t)</math> gets us <math>f(t)=u-3u^2</math>. Since the vertex of a quadratic is <math>\frac{-b}{2a}</math>, the <math>u</math> value of the vertex is <math>\frac{-1}{2(-3)}=\frac{1}{6}</math>. From there, we can obtain the maximum value of <math>f(t)</math>, which is <math>(\frac{1}{6})-3(\frac{1}{6})^2=\frac{1}{12}</math>, so the answer is <math>\boxed{C=\frac{1}{12}}</math>. | ||
+ | |||
+ | ~ Randomlygenerated | ||
− | |||
==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 |
Latest revision as of 23:24, 24 October 2021
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 . Rembering 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 (Simple)
Set , now we get and we want to find the maximum value of that expression. And that expression can be further simplified to , now lets set as and we get . Expand and we get . We can make it simpler by rearranging and getting as the original expression, so, now we need to find the minimum value of and take the negative of that and get our answer. So, we complete the square on to get it as . And by the trivial inequality, any real number squared is at least , so, to minimize this expression, we just set the square as and you subtract from to get as the minimum value of the expression and you take the negative of that to get as the final answer.
~ math31415926535
Solution 7 (Simpler)
Distribute the terms and simplify to get . From there, we can set , and substituting into gets us . Since the vertex of a quadratic is , the value of the vertex is . From there, we can obtain the maximum value of , which is , so the answer is .
~ Randomlygenerated
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 |
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