anonymous
  • anonymous
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.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • 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.
anonymous
  • anonymous
please me how
anonymous
  • anonymous
Do you know calculus?

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anonymous
  • anonymous
precalculus
anonymous
  • 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.

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