Question

If x and y are two real numbers such that x > 0 and xy = 1. The the minimum value of x + y is ________________.

Answer 1

It is given that, x and y are two real numbers such that x > 0 and xy = 1.

Let S = x + y.

Now, $xy=1\Rightarrow y=\frac{1}{x}$

$\therefore S=x+y=x+\frac{1}{x}$

Differentiating both sides with respect to x, we get

$\frac{dS}{dx}=1-\frac{1}{x^2}$

For maxima or minima,

$\frac{dS}{dx}=0$

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

$\Rightarrow x^2=1$

⇒ x = 1 (x > 0)

Now,

$\frac{d^{2}S}{d{x}^2}=\frac{2}{x^3}$

At x = 1, we have

${\left(\frac{d^{2}S}{d{x}^2}\right)}_{x=1}=\frac{2}{{\left(1\right)}^3}=2>0$

So, x = 1 is the point of local minimum.

Thus, S is minimum when x = 1.

When x = 1, $y=\frac{1}{x}=1$

∴ Minimum value of S = x + y = 1 + 1 = 2

Thus, the minimum value of x + y is 2.


If x and y are two real numbers such that x > 0 and xy = 1. The the minimum value of x + y is ___2___.