1983 AIME Problems/Problem 9
Find the minimum value of for .
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 plugging in for in the original equation (when ; since is continuous and increasing on the interval and its range on that interval is from , by the Intermediate Value Theorem this value is attainable).
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 .
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 of 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 0. Simplifying, we have and . Since both and are positive on the given interval, we can ignore the negative result. Plugging into our expression for , we have .
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