Difference between revisions of "1983 AIME Problems/Problem 2"
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Edit: <math>|x-p-15|</math> can equal <math>15+p-x</math> or <math>x-p-15</math> (for example, if <math>x=7</math> and <math>p=-12</math>, <math>x-p-15=4</math>). Thus, our two "cases" are | Edit: <math>|x-p-15|</math> can equal <math>15+p-x</math> or <math>x-p-15</math> (for example, if <math>x=7</math> and <math>p=-12</math>, <math>x-p-15=4</math>). Thus, our two "cases" are | ||
| − | <math>30-x</math> (if <math>x-p\leq15</math>) and <math>x-2p</math> (if <math>x-p\geq15</math>). However, both of these cases give us <math>\boxed{ | + | <math>30-x</math> (if <math>x-p\leq15</math>) and <math>x-2p</math> (if <math>x-p\geq15</math>). However, both of these cases give us <math>\boxed{015}</math> as the minimum value for <math>f(x)</math>, which indeed is the answer posted above. |
Revision as of 18:20, 2 October 2016
Problem
Let 
, where 
. Determine the minimum value taken by 
for 
in the interval 
.
Solution
It is best to get rid of the absolute value first.
Under the given circumstances, we notice that 
, 
, and 
.
Adding these together, we find that the sum is equal to 
, of which the minimum value is attained when 
.
Edit: 
can equal 
or 
(for example, if 
and 
, 
). Thus, our two "cases" are
(if 
) and 
(if 
). However, both of these cases give us 
as the minimum value for , which indeed is the answer posted above.
Also note the lowest value occurs when 
because this make the first two requirements 
. It is easy then to check that 15 is the minimum value.
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 | ||
