Difference between revisions of "2000 AMC 12 Problems/Problem 12"
m (→Solution: typo) |
m (rigorize) |
||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | Let A, M, and C be nonnegative | + | 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 == | ||
− | + | <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) + (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>. | |
− | + | <cmath>A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C = 125 - 13 = 112 \Rightarrow \mathrm{(E)}</cmath> | |
− | |||
== See also == | == See also == |
Revision as of 17:44, 4 January 2008
Problem
Let and be nonnegative integers such that . What is the maximum value of ?
Solution
By AM-GM, ; thus is maximized at , which occurs when .
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 |