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>. Rembering what we want to find(divide by <math>4^t</math>), we get the maximal values as <math>\frac{1}{12}</math>, and we are done. | + | 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(divide by <math>4^t</math>), we get the maximal values as <math>\frac{1}{12}</math>, and we are done. |
==Solution 2== | ==Solution 2== |
Revision as of 02:39, 28 July 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 .
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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