1983 AIME Problems/Problem 2: Difference between revisions
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<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{15}</math> as the minimum value for <math>f(x)</math>, which indeed is the answer posted above. | <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{15}</math> as the minimum value for <math>f(x)</math>, which indeed is the answer posted above. | ||
== See | == See Also == | ||
{{AIME box|year=1983|num-b=1|num-a=3}} | {{AIME box|year=1983|num-b=1|num-a=3}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 05:59, 16 April 2012
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.
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 | ||
