Difference between revisions of "1965 AHSME Problems/Problem 34"
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− | After this simplification, we may notice that we may use calculus, or the AM-GM inequality to finish this problem because <math>x\ge 0</math>, which implies that both <math>\frac{2(x+1)}{3} \text{ and } \frac{3}{2(x+1)}</math> are greater than zero. Continuing with AM-GM: | + | After this simplification, we may notice that we may use calculus, or the [[AM-GM inequality]] to finish this problem because <math>x\ge 0</math>, which implies that both <math>\frac{2(x+1)}{3} \text{ and } \frac{3}{2(x+1)}</math> are greater than zero. Continuing with AM-GM: |
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Revision as of 11:40, 19 July 2024
Contents
[hide]Problem 34
For the smallest value of is:
Solution 1
To begin, lets denote the equation, as . Let's notice that:
After this simplification, we may notice that we may use calculus, or the AM-GM inequality to finish this problem because , which implies that both are greater than zero. Continuing with AM-GM:
Therefore, ,
Solution 2 (Calculus)
Let .
Take the derivative of using the quotient rule.
Thus, our answer is .
Solution 3 (answer choices, no AM-GM or calculus)
We go from A through E and we look to find the smallest value so that , so we start from A:
However by the quadratic formula there are no real solutions of , so cannot be greater than 0. We move on to B:
There is one solution: , which is greater than 0, so 2 works as a value. Since all the other options are bigger than 2 or invalid, the answer must be