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== Problem ==
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>

== Solution ==
<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\\
&= A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + 13\end{align*}</cmath>

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

<cmath>A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C = 125 - 13 = 112 \Rightarrow \mathrm{(E)}</cmath>

Alternatively, a quick AMC heuristic method would be to consider the equality case since that often gives a maximum or minimum. So <math>A=M=C=4</math> gives <math>AMC+AM+AC+MC = 112</math>, and that is the greatest answer choice, so the answer is <math>\mathrm{E}</math>.

== See also ==
{{AMC12 box|year=2000|num-b=11|num-a=13}}

[[Category:Introductory Algebra Problems]]