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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.

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  1. anonymous
    • 5 years ago
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    \[ A + B = 9 \text {, for } A > 0 \text { and } B > 0 \] \[ \rightarrow A = 9-B \] 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
    • 5 years ago
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    please me how

  3. anonymous
    • 5 years ago
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    Do you know calculus?

  4. anonymous
    • 5 years ago
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    precalculus

  5. anonymous
    • 5 years ago
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    \[ S = \frac{1}{9-B} + \frac{1}{B} \] \[ = \frac{B}{B(9-B)} + \frac{(9-B)}{B(9-B)} \] \[ = \frac{9}{B(9-B)} = \frac{9}{9B - B^2}\] 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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