Difference between revisions of "1983 AIME Problems/Problem 2"
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Adding these together, we find that the sum is equal to <math>30-x</math>, of which the minimum value is attained when <math>x=\boxed{015}</math>. | Adding these together, we find that the sum is equal to <math>30-x</math>, of which the minimum value is attained when <math>x=\boxed{015}</math>. | ||
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Also note the lowest value occurs when <math>x=p=15</math> because this make the first two requirements <math>0</math>. It is easy then to check that 15 is the minimum value. | Also note the lowest value occurs when <math>x=p=15</math> because this make the first two requirements <math>0</math>. It is easy then to check that 15 is the minimum value. |
Revision as of 00:56, 2 July 2018
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
.
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 |