Question

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.

Answer 1

\[ A + B = 9 \text {, for } A > 0 \text { and } B > 0 \]
\[ \rightarrow A = 9-B \]
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.

Answer 2

please me how

Answer 3

Do you know calculus?

Answer 4

precalculus

Answer 5

\[ S = \frac{1}{9-B} + \frac{1}{B} \]
\[ = \frac{B}{B(9-B)} + \frac{(9-B)}{B(9-B)} \]
\[ = \frac{9}{B(9-B)} = \frac{9}{9B - B^2}\]

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.