Question

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

• A. $1$
• B. $2$
• C. $2\left(\frac{1}{2}\right)$
• D. $3\left(\frac{1}{3}\right)$

Answer 1

The correct option is B

$2$

Explanation for the correct answer:

Finding the minimum value of $x+y$ :

Given,

$$\begin{gathered} xy=1 \\ y=\frac{1}{x} \end{gathered}$$

Then,

$\begin{array}{rcl}s& =& x+y\\ & =& x+\frac{1}{x}\end{array}$

Differentiate the above equation with respect to $x$.

$\begin{array}{rcl}\frac{ds}{dx}& =& [1–\left(\frac{1}{x^2}\right)]...\left(1\right)\\ & =& \frac{[x^2–1]}{x^2}\end{array}$

Now, differentiate $\left(1\right)$ with respect to $x$.

$\begin{array}{rcl}\frac{d^{2}s}{d{x}^2}& =& \frac{2}{x^3}\end{array}$

For minimum,

$$\begin{gathered} \frac{ds}{dx}=0 \\ \Rightarrow x^2–1=0 \\ \Rightarrow x=\pm 1 \\ {\left(\frac{d^{2}s}{d{x}^2}\right)}_{x=-1}=-2<0and \\ {\left(\frac{d^{2}s}{d{x}^2}\right)}_{x=1}=2>0 \end{gathered}$$

Therefore, $x+y$ is minimum at $x=1$ and minimum value of $x+y=(1+1)=2$

Hence, the correct is option B