Question

The sum of two non-zero numbers is 8, the minimum value of the sum of the reciprocals is

(a) $\frac{1}{4}$

(b) $\frac{1}{2}$

(c) $\frac{1}{8}$

(d) none of these

Answer 1

$$\begin{gathered} \left(b\right)\frac{1}{2} \\ \mathrm{Let}\mathrm{the}\mathrm{two}\mathrm{non}-\mathrm{zero}\mathrm{numbers}\mathrm{be}\mathrm{x}\mathrm{and}\mathrm{y}.\mathrm{Then}, \\ x+y=8 \\ \Rightarrow y=8-x...\left(1\right) \\ \mathrm{Now}, \\ f\left(x\right)=\frac{1}{x}+\frac{1}{y} \\ \Rightarrow f\left(x\right)=\frac{1}{x}+\frac{1}{8-x}\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow f\text{'}\left(x\right)=\frac{-1}{x^2}+\frac{1}{{\left(8-x\right)}^2} \\ \text{For a local minima or a local maxima, we must have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow \frac{-1}{x^2}+\frac{1}{{\left(8-x\right)}^2}=0 \\ \Rightarrow \frac{-{\left(8-x\right)}^2+x^2}{{\left(x\right)}^2{\left(8-x\right)}^2}=0 \\ \Rightarrow -64-x^2+16x+x^2=0 \\ \Rightarrow 16x-64=0 \\ \Rightarrow x=4 \\ f\text{'}\text{'}\left(x\right)=\frac{2}{x^3}-\frac{2}{{\left(8-x\right)}^3} \\ \Rightarrow f\text{'}\text{'}\left(4\right)=\frac{2}{4^3}-\frac{2}{{\left(8-4\right)}^3} \\ \Rightarrow f\text{'}\text{'}\left(4\right)=\frac{2}{64}-\frac{2}{64}=0 \\ \therefore \text{Minimum value}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \end{gathered}$$