## anonymous 5 years ago The sum of two positive integers is 9. What is the least possible sum of their reciprocals? Express your answer as a decimal to the nearest hundredth? show all work.

1. anonymous

$A + B = 9 \text {, for } A > 0 \text { and } B > 0$ $\rightarrow A = 9-B$ The sum of their reciprocals is $S = \frac{1}{A} + \frac{1}{B} = \frac{1}{9-B} + \frac{1}{B}$ Find the minimum value for S.

2. anonymous

3. anonymous

Do you know calculus?

4. anonymous

precalculus

5. anonymous

$S = \frac{1}{9-B} + \frac{1}{B}$ $= \frac{B}{B(9-B)} + \frac{(9-B)}{B(9-B)}$ $= \frac{9}{B(9-B)} = \frac{9}{9B - B^2}$ Now, since the numerator is constant the fraction will be at its smallest value when the denominator is at the largest value. So what is the largest possible value for $-B^2 + 9B + 0$ With B restricted to being > 0.