Question

If sum of two numbers is $6$, then the minimum value of the sum of their reciprocals is

• A. $\frac{6}{5}$
• B. $\frac{3}{4}$
• C. $\frac{2}{3}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C

$\frac{2}{3}$

Explanation for the correct option:

Step1:Let the two numbers be $a$ and $b$.

$\begin{array}{c}a+b=6\\ b=6-a\\ s=\left(\frac{1}{a}\right)+\left(\frac{1}{b}\right)\\ =\left(\frac{1}{a}\right)+\left(\frac{1}{6-a}\right)\end{array}$

Step 2: Differentiate the first and second derivative

$\begin{array}{c}\frac{ds}{da}=\left(\frac{-1}{a^2}\right)+\left(\frac{-1}{{\left(6-a\right)}^2}\right)\times \left(-1\right)\\ =0\end{array}$

$\begin{array}{l}\Rightarrow \frac{-1}{a^2}=\frac{-1}{\left(6-a^2\right)}\\ \Rightarrow a^2={\left(6-a\right)}^2\\ \Rightarrow a=6-a\\ \Rightarrow a=3\end{array}$

Find the second derivative

$\begin{array}{c}\therefore \frac{d^{2}s}{d{a}^2}=\frac{2}{a^3}-\left(\frac{2}{{\left(6-a\right)}^3}\right)\times \left(-1\right)\\ =\left(\frac{2}{3^3}\right)+\left(\frac{2}{3^3}\right)>0\\ s=\left(\frac{1}{3}\right)+\left(\frac{1}{3}\right)\\ =\frac{2}{3}\end{array}$

Hence option (3) is the answer