1983 AIME Problems/Problem 2
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 |