Difference between revisions of "2000 AMC 12 Problems/Problem 12"
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<math> \mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D) \ 102 } \qquad \mathrm{(E) \ 112 } </math> | <math> \mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D) \ 102 } \qquad \mathrm{(E) \ 112 } </math> | ||
− | == Solution == | + | == Solution 1 == |
It is not hard to see that | It is not hard to see that | ||
<cmath>(A+1)(M+1)(C+1)=</cmath> | <cmath>(A+1)(M+1)(C+1)=</cmath> | ||
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so the answer is <math>\boxed{\text{E}}</math>. | so the answer is <math>\boxed{\text{E}}</math>. | ||
− | + | == Solution 2 == | |
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 shape make it a square) then you can try to make <math>A,M</math> and <math>C</math> as close as possible. In this case, they would all be equal to <math>4</math>, so <math>AMC+AM+AC+MC=64+16+16+16=112</math>, giving you the answer of <math>\boxed{\text{E}}</math>. | 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 shape make it a square) then you can try to make <math>A,M</math> and <math>C</math> as close as possible. In this case, they would all be equal to <math>4</math>, so <math>AMC+AM+AC+MC=64+16+16+16=112</math>, giving you the answer of <math>\boxed{\text{E}}</math>. |
Revision as of 17:26, 28 August 2017
Contents
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
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 shape 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 , giving you the answer of .
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.