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Difference between revisions of "2000 AMC 12 Problems/Problem 12"

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== Problem ==
 
== Problem ==
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Let A, M, and C be nonnegative integers such that <math>\displaystyle A + M + C=12</math>. What is the maximum value of <math>A \cdot M \cdot C</math>+<math>A \cdot M</math>+<math>M \cdot C</math>+<math>A\cdot C</math>?
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<!-- 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>
  
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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>
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[katex] \mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D) \ 102 } \qquad \mathrm{(E) \ 112 }  [/katex]
  
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== Solution ==
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== Solution 1 ==
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:<math>(A + 1)(M + 1)(C + 1) = A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + A + M + C + 1</math>
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It is not hard to see that
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:<math>(A + 1)(M + 1)(C + 1) = A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + 13</math>
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<cmath>(A+1)(M+1)(C+1)=</cmath>
 +
<cmath>AMC+AM+AC+MC+A+M+C+1</cmath>
 +
Since <math>A+M+C=12</math>, we can rewrite this as
 +
<cmath>(A+1)(M+1)(C+1)=</cmath>
 +
<cmath>AMC+AM+AC+MC+13</cmath>
 +
So we wish to maximize
 +
<cmath>(A+1)(M+1)(C+1)-13</cmath>
 +
Which is largest when all the factors are equal (consequence of AM-GM).  Since <math>A+M+C=12</math>, we set <math>A=M=C=4</math>
 +
Which gives us
 +
<cmath>(4+1)(4+1)(4+1)-13=112</cmath>
 +
so the answer is <math>\boxed{\textbf{(E) }112}.</math>
  
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The [[term]] <math>(A + 1)(M + 1)(C + 1)</math> is [[maximum|maximized]] when A, M, and C are closed together, which in this case would be if all of them were 4. Thus,
+
== Solution 2 (Nonrigorous) ==
 +
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>.
  
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:<math>125 = A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + 13</math>
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== Solution 3 (Answer Choices) ==
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:<math>A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C = 112 \Rightarrow \mathrm{E}</math>
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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>.
 +
 
 +
== Solution 4 (Semi-rigorous) ==
 +
Given that <math>A</math>, <math>M</math>, and <math>C</math> are nonnegative integers, it should be intuitive that maximizing <math>AMC</math> maximizes <math>AM + MC + CA</math>. We thus only need to maximize <math>AMC</math>. By the [[AM-GM Inequality]], <cmath>\frac{A + M + C}{3} \geq \sqrt[3]{AMC},</cmath> with equality if and only if <math>A = M = C</math>. Note that the maximum of <math>AMC</math> occurs under the equality condition --- hence, all three variables are equal. The rest of the problem is smooth sailing; <math>A + M + C = 12</math> implies that <math>A = M = C = 4</math>, so <math>AMC + AM + MC + CA = 4^3 + 3*4^2 = 112.</math> The answer is thus <math>\boxed{\text{E}}</math>, as required.
 +
 
 +
== Solution 5 (Double AM-GM) ==
 +
We start off the same way as Solution 4, using AM-GM to observe that <math>AMC \leq 64</math>. We then observe that
 +
 
 +
<math>(A + M + C)^2 = A^2 + M^2 + C^2 + 2(AM + MC + AC) = 144</math>, since <math>A + M + C = 12</math>.
 +
 
 +
We can use the AM-GM inequality again, this time observing that <math>\frac{A^2 + M^2 + C^2}{3} \geq \sqrt[3]{{(AMC)}^2}</math>
 +
 
 +
Since <math>AMC \leq 64</math>,  <math>3 \sqrt[3]{{(AMC)}^2} \leq 48</math>. We then plug this in to yield
 +
 
 +
<math>A^2 + M^2 + C^2 + 2(AM + MC + AC) = 144 \geq 48 + 2(AM + MC + AC)</math>
 +
 
 +
Thus, <math>AM + MC + AC \leq 48</math>. We now revisit the original equation that we wish to maximize. Since we know <math>AMC \leq 64</math>, we now have upper bounds on both of our unruly terms. Plugging both in results in <math>48 + 64 = \boxed{\textbf{(E) }112}</math>
 +
 
 +
== 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
 +
 
 +
== Video Solution ==
 +
https://youtu.be/lxqxQhGterg
  
 
== See also ==
 
== See also ==
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*[[2000 AMC 12]]
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{{AMC12 box|year=2000|num-b=11|num-a=13}}
  
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{{AMC12 box|year=2000|num-b=11|num-a=13}}
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[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 15:15, 27 July 2024

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 \[(A+1)(M+1)(C+1)=\] \[AMC+AM+AC+MC+A+M+C+1\] Since $A+M+C=12$, we can rewrite this as \[(A+1)(M+1)(C+1)=\] \[AMC+AM+AC+MC+13\] So we wish to maximize \[(A+1)(M+1)(C+1)-13\] Which is largest when all the factors are equal (consequence of AM-GM). Since $A+M+C=12$, we set $A=M=C=4$ Which gives us \[(4+1)(4+1)(4+1)-13=112\] so the answer is $\boxed{\textbf{(E) }112}.$

Solution 2 (Nonrigorous)

If you know that to maximize your result you $\textit{usually}$ 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 $A,M$ and $C$ as close as possible. In this case, they would all be equal to $4$, so $AMC+AM+AC+MC=64+16+16+16=112$, $\boxed{\text{E}}$.

Solution 3 (Answer Choices)

Assume $A$, $M$, and $C$ are equal to $4$. Since the resulting value of $AMC+AM+AC+MC$ will be $112$ and this is the largest answer choice, our answer is $\boxed{\textbf{(E) }112}$.

Solution 4 (Semi-rigorous)

Given that $A$, $M$, and $C$ are nonnegative integers, it should be intuitive that maximizing $AMC$ maximizes $AM + MC + CA$. We thus only need to maximize $AMC$. By the AM-GM Inequality, \[\frac{A + M + C}{3} \geq \sqrt[3]{AMC},\] with equality if and only if $A = M = C$. Note that the maximum of $AMC$ occurs under the equality condition --- hence, all three variables are equal. The rest of the problem is smooth sailing; $A + M + C = 12$ implies that $A = M = C = 4$, so $AMC + AM + MC + CA = 4^3 + 3*4^2 = 112.$ The answer is thus $\boxed{\text{E}}$, as required.

Solution 5 (Double AM-GM)

We start off the same way as Solution 4, using AM-GM to observe that $AMC \leq 64$. We then observe that

$(A + M + C)^2 = A^2 + M^2 + C^2 + 2(AM + MC + AC) = 144$, since $A + M + C = 12$.

We can use the AM-GM inequality again, this time observing that $\frac{A^2 + M^2 + C^2}{3} \geq \sqrt[3]{{(AMC)}^2}$

Since $AMC \leq 64$, $3 \sqrt[3]{{(AMC)}^2} \leq 48$. We then plug this in to yield

$A^2 + M^2 + C^2 + 2(AM + MC + AC) = 144 \geq 48 + 2(AM + MC + AC)$

Thus, $AM + MC + AC \leq 48$. We now revisit the original equation that we wish to maximize. Since we know $AMC \leq 64$, we now have upper bounds on both of our unruly terms. Plugging both in results in $48 + 64 = \boxed{\textbf{(E) }112}$

Solution 6 (Optimization)

The largest number for our value would be $A = M = C.$ So $3A = 12$ and $A = M = C = 4.$ $4\times4\times4 + 4\times4 + 4\times4 = 112$ or $\boxed{\textbf{(E) }112}$

~BowenNa

Video Solution

https://youtu.be/lxqxQhGterg

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

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