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>. 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. | + | 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== |
Latest revision as of 19:09, 13 October 2024
Contents
[hide]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.