1983 AIME Problems/Problem 9
Contents
Problem
Find the minimum value of for
.
Solution 1
Let . We can rewrite the expression as
.
Since , and
because
, we have
. So we can apply AM-GM:
The equality holds when .
Therefore, the minimum value is . This is reached when we have
in the original equation (since
is continuous and increasing on the interval
, and its range on that interval is from
, this value of
is attainable by the Intermediate Value Theorem).
Solution 2
We can rewrite the numerator to be a perfect square by adding . Thus, we must also add back
.
This results in .
Thus, if , then the minimum is obviously
. We show this possible with the same methods in Solution 1; thus the answer is
.
Solution 3 (uses calculus)
Let and rewrite the expression as
, similar to the previous solution. To minimize
, take the derivative of
and set it equal to zero.
The derivative of , using the Power Rule, is
=
is zero only when
or
. It can further be verified that
and
are relative minima by finding the derivatives at other points near the critical points. However, since
is always positive in the given domain,
. Therefore,
=
, and the answer is
.
Solution 4 (also uses calculus)
As above, let . Add
to the expression and subtract
, giving
. Taking the derivative of
using the Chain Rule and Quotient Rule, we have
. We find the minimum value by setting this to
. Simplifying, we have
and
. Since both
and
are positive on the given interval, we can ignore the negative root. Plugging
into our expression for
, we have
.
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |