Difference between revisions of "2020 AMC 12B Problems/Problem 22"
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==Video Solution== | ==Video Solution== |
Revision as of 17:18, 28 August 2022
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
. 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 Solution1
~Education, the Study of Everything
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.