2000 AMC 12 Problems/Problem 12
Contents
Problem
Let [mathjax]A, M,[/mathjax] and [mathjax]C[/mathjax] be nonnegative integers such that [mathjax]A + M + C=12[/mathjax]. What is the maximum value of [mathjax]A \cdot M \cdot C + A \cdot M + M \cdot C + A \cdot C[/mathjax]?
[katex] \mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D) \ 102 } \qquad \mathrm{(E) \ 112 } [/katex]
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 (Answer Choices)
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
Solution 6 (Optimization)
The largest number for our value would be So and or
~BowenNa
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.