Difference between revisions of "1980 AHSME Problems/Problem 22"
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== Solution == | == Solution == | ||
| − | <math>\ | + | The first two given functions intersect at <math>\left(\frac{1}{3},\frac{7}{3}\right)</math>, and last two at <math>\left(\frac{2}{3},\frac{8}{3}\right)</math>. Therefore <cmath>f(x)=\left\{ \begin{matrix} 4x+1 & x<\frac{1}{3} \\ |
| + | x+2 & \frac{1}{3}>x>\frac{2}{3} \\ | ||
| + | -2x+4 & x>\frac{2}{3} | ||
| + | \end{matrix}\right. | ||
| + | </cmath> | ||
| + | Which attains a maximum at | ||
| + | <math>\boxed{(E)\ \frac{8}{3}}</math> | ||
== See also == | == See also == | ||
Latest revision as of 23:32, 22 July 2016
Problem
For each real number 
, let 
be the minimum of the numbers 
, and 
. Then the maximum value of is
Solution
The first two given functions intersect at 
, and last two at 
. Therefore ![\[f(x)=\left\{ \begin{matrix} 4x+1 & x<\frac{1}{3} \\ x+2 & \frac{1}{3}>x>\frac{2}{3} \\ -2x+4 & x>\frac{2}{3} \end{matrix}\right.\]](http://latex.artofproblemsolving.com/e/a/7/ea7cbf06a56fe7bd22f8b58a6a64ac8ad0382c7d.png)
Which attains a maximum at
See also
| 1980 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
