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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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Let <math>A, M,</math> and <math>C</math> be [[nonnegative integer]]s such that <math>A + M + C=12</math>. What is the maximum value of <math>A \cdot M \cdot C + A \cdot M + M \cdot C + A \cdot C</math>?
  
 
<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 ==
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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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<cmath>\begin{align*}(A + 1)(M + 1)(C + 1) &= A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + A + M + C + 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 + 13</math>
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&= A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + 13\end{align*}</cmath>
  
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The [[term]] <math>(A + 1)(M + 1)(C + 1)</math> is [[maximum|maximized]] when A, M, and C are close together, which in this case would be if all of them were 4. Thus,
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By [[AM-GM]], <math>\frac{(A+1) + (B+1) + (C+1)}{3} = 5 \ge \sqrt[3]{(A+1)(B+1)(C+1)}</math>; thus <math>(A+1)(B+1)(C+1)</math> is [[maximum|maximized]] at <math>125</math>, which occurs when <math>A=B=C=4</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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<cmath>A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C = 125 - 13 = 112 \Rightarrow \mathrm{(E)}</cmath>
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:<math>A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C = 112 \Rightarrow \mathrm{E}</math>
 
  
 
== See also ==
 
== See also ==

Revision as of 18:44, 4 January 2008

Problem

Let $A, M,$ and $C$ be nonnegative integers such that $A + M + C=12$. What is the maximum value of $A \cdot M \cdot C + A \cdot M + M \cdot C + A \cdot C$?

$\mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D) \ 102 } \qquad \mathrm{(E) \ 112 }$

Solution

\begin{align*}(A + 1)(M + 1)(C + 1) &= A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + A + M + C + 1\\ &= A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + 13\end{align*}

By AM-GM, $\frac{(A+1) + (B+1) + (C+1)}{3} = 5 \ge \sqrt[3]{(A+1)(B+1)(C+1)}$; thus $(A+1)(B+1)(C+1)$ is maximized at $125$, which occurs when $A=B=C=4$.

\[A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C = 125 - 13 = 112 \Rightarrow \mathrm{(E)}\]

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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