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

Question

KollegeApply · Accepted Answer

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

AChoose Your Answer