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 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).
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, or by checking that the second derivative is positive. 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 .
Set equal to . Then multiply by on both sides to get . We then subtract from both sides to get . This looks like a quadratic so set and use quadratic equation on to see that . We know that must be an integer and as small as it can be, so = 12. We plug this back in to see that which we can prove works using methods from solution 1. This makes the answer
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