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 |