Question

The sum of two non-zero numbers is 6. The minimum value of the sum of their reciprocals is ______.

Options

• ${\displaystyle \frac{3}{4}}$

• ${\displaystyle \frac{6}{5}}$

• ${\displaystyle \frac{2}{3}}$

• None of these

Answer 1

The sum of two non-zero numbers is 6. The minimum value of the sum of their reciprocals is ${\displaystyle \frac{2}{3}}$.

Explanation:

Let x + y = 6

⇒ y = 6 – x

${\displaystyle \frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}}$

f(x) = ${\displaystyle \frac{6}{xy}=\frac{6}{x(6-x)}=\frac{6}{6x-x^2}}$

∴ f'(x) = ${\displaystyle \frac{-6}{{(6x-x^2)}^2}\cdot (6-2x)}$

For maximum or minimum of f(x),

f'(x) = 0

⇒ 6 – 2x = 0

∴ x = 3 and y = 3

∴ ${\displaystyle \text{min}(\frac{1}{x}+\frac{1}{y})=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}}$