Difference between revisions of "2000 AMC 12 Problems/Problem 12"
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== Problem == | == Problem == | ||
− | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let | + | <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let [mathjax]A, M,[/mathjax] and [mathjax]C[/mathjax] be [[nonnegative integer]]s 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]?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
− | + | [katex] \mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D) \ 102 } \qquad \mathrm{(E) \ 112 } [/katex] | |
== Solution 1 == | == Solution 1 == | ||
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If you know that to maximize your result you <math>\textit{usually}</math> 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 <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>, <math>\boxed{\text{E}}</math>. | If you know that to maximize your result you <math>\textit{usually}</math> 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 <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>, <math>\boxed{\text{E}}</math>. | ||
− | == Solution 3 == | + | == Solution 3 (Answer Choices) == |
Assume <math>A</math>, <math>M</math>, and <math>C</math> are equal to <math>4</math>. Since the resulting value of <math>AMC+AM+AC+MC</math> will be <math>112</math> and this is the largest answer choice, our answer is <math>\boxed{\textbf{(E) }112}</math>. | Assume <math>A</math>, <math>M</math>, and <math>C</math> are equal to <math>4</math>. Since the resulting value of <math>AMC+AM+AC+MC</math> will be <math>112</math> and this is the largest answer choice, our answer is <math>\boxed{\textbf{(E) }112}</math>. | ||
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== Solution 6 (Optimization) == | == Solution 6 (Optimization) == | ||
− | + | The largest number for our value would be <math>A = M = C.</math> So <math>3A = 12 </math> and <math>A = M = C = 4.</math> <math>4\times4\times4 + 4\times4 + 4\times4 = 112 </math> or <math>\boxed{\textbf{(E) }112}</math> | |
~BowenNa | ~BowenNa |
Latest revision as of 16:15, 27 July 2024
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.