Difference between revisions of "2020 AMC 12B Problems/Problem 22"
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==Solution 3== | ==Solution 3== | ||
Take the derivative of this function and let the derivative equals to 0, then this gives you <math>2^t=6t</math>. Substitute it into the original function you can get <math>\boxed{C}</math>. | Take the derivative of this function and let the derivative equals to 0, then this gives you <math>2^t=6t</math>. Substitute it into the original function you can get <math>\boxed{C}</math>. | ||
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+ | ==See Also== | ||
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+ | {{AMC12 box|year=2020|ab=B|num-b=15|num-a=17}} | ||
+ | {{MAA Notice}} |
Revision as of 02:30, 8 February 2020
Problem 22
What is the maximum value of for real values of
Solution1
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
.
Solution2
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 that 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 3
Take the derivative of this function and let the derivative equals to 0, then this gives you . Substitute it into the original function you can get
.
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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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