Question

Let $x$ and $y$ be two positive real numbers such that $xy=1.$ The minimum value of $x+y$ is

• A. $1$
• B. $1/2$
• C. $2$
• D. $1/4$

Answer 1

The correct option is C $2$

Given $xy=1$ and $f(x,y)=x+y$
$\Rightarrow f\left(x\right)=x+\frac{1}{x}$
$f^{\prime }\left(x\right)=1-\frac{1}{x^2}$
For maxima or minima,
$f^{\prime }\left(x\right)=0$
$\Rightarrow x=\pm 1$
$f^{\prime \prime }\left(x\right)=\frac{2}{x^3}$
$f^{\prime \prime }\left(x\right)>0$ at $x=1$
Hence f(x) has minimum at $x=1$
$f\left(1\right)=2$
So, minimum value of $x+yis2$.