Difference between revisions of "2000 AMC 12 Problems/Problem 12"
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Which gives us | Which gives us | ||
<cmath>(4+1)(4+1)(4+1)-13=112</cmath> | <cmath>(4+1)(4+1)(4+1)-13=112</cmath> | ||
− | so the answer is \boxed{\textbf{(E) }112}. | + | so the answer is <math>\boxed{\textbf{(E) }112}.</math> |
== Solution 2 (Nonrigorous) == | == Solution 2 (Nonrigorous) == |
Revision as of 02:02, 30 July 2023
Contents
[hide]Problem
Let and be nonnegative integers such that . What is the maximum value of ?
Solution 1
It is not hard to see that Since , we can rewrite this as So we wish to maximize Which is largest when all the factors are equal (consequence of AM-GM). Since , we set Which gives us so the answer is
Solution 2 (Nonrigorous)
If you know that to maximize your result you have to make the numbers as close together as possible, (for example to maximize area for a polygon make it a square) then you can try to make and as close as possible. In this case, they would all be equal to , so , .
Solution 3
Assume , , and are equal to . Since the resulting value of will be and this is the largest answer choice, our answer is .
Solution 4 (Semi-rigorous)
Given that , , and are nonnegative integers, it should be intuitive that maximizing maximizes . We thus only need to maximize . By the AM-GM Inequality, with equality if and only if . Note that the maximum of occurs under the equality condition --- hence, all three variables are equal. The rest of the problem is smooth sailing; implies that , so The answer is thus , as required.
Solution 5 (Double AM-GM)
We start off the same way as Solution 4, using AM-GM to observe that . We then observe that
, since .
We can use the AM-GM inequality again, this time observing that
Since , . We then plug this in to yield
Thus, . We now revisit the original equation that we wish to maximize. Since we know , we now have upper bounds on both of our unruly terms. Plugging both in results in
Video Solution
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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.