Question

Let x and y be two real numbers such that x > 0 and 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

Let $z=x+y=x+1/x$, since $(xy=1)$
$\frac{dz}{dx}=1-1/x^2$
For minimum value of $z$
$\frac{dz}{dx}=0=1-1/x^2\Rightarrow x=1$, $since(x>0)$
Therefore minimum value of $z=2$