#### Question

Let x, y be two variables and x>0, xy=1, then minimum value of x+y is

(a) 1

(b) 2

(c) \[2\frac{1}{2}\]

(d) \[3\frac{1}{3}\]

#### Solution

(b) 2

\[\text { Given }: xy = 1\]

\[ \Rightarrow y = \frac{1}{x}\]

\[f\left( x \right) = x + \frac{1}{x}\]

\[ \Rightarrow f'\left( x \right) = 1 - \frac{1}{x^2}\]

\[\text { For a local maxima or a local minima, we must have } \]

\[f'\left( x \right) = 0\]

\[ \Rightarrow 1 - \frac{1}{x^2} = 0\]

\[ \Rightarrow x^2 - 1 = 0\]

\[ \Rightarrow x^2 = 1\]

\[ \Rightarrow x = \pm 1\]

\[ \Rightarrow x = 1 \left( \text { Given }: x > 1 \right)\]

\[ \Rightarro

\[\text { Now,} \]

\[f''\left( x \right) = \frac{2}{x^3}\]

\[ \Rightarrow f''\left( 1 \right) = 2 > 0\]

\[\text { So, x = 1 is a local minima } . \]

\[ \therefore \text { Minimum value of } f\left( x \right) = f\left( 1 \right) = 1 + 1 = 2\]