Difference between revisions of "1983 AIME Problems/Problem 2"
Speedstorm (talk | contribs) m (→Solution) |
Hashtagmath (talk | contribs) m |
||
| Line 1: | Line 1: | ||
| + | __TOC__ | ||
| + | |||
== Problem == | == Problem == | ||
Let <math>f(x)=|x-p|+|x-15|+|x-p-15|</math>, where <math>0 < p < 15</math>. Determine the [[minimum]] value taken by <math>f(x)</math> for <math>x</math> in the [[interval]] <math>p \leq x\leq15</math>. | Let <math>f(x)=|x-p|+|x-15|+|x-p-15|</math>, where <math>0 < p < 15</math>. Determine the [[minimum]] value taken by <math>f(x)</math> for <math>x</math> in the [[interval]] <math>p \leq x\leq15</math>. | ||
Revision as of 15:55, 24 December 2022
Contents
Problem
Let 
, where 
. Determine the minimum value taken by 
for 
in the interval 
.
Solution
It is best to get rid of the absolute values first.
Under the given circumstances, we notice that 
, 
, and 
.
Adding these together, we find that the sum is equal to 
, which attains its minimum value (on the given interval 
) when 
, giving a minimum of 
.
See Also
| 1983 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
