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

1.	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. \(3\frac{1}{3}\)
Answer» Option : (B) xy = 1, x > 0, y > 0 ⇒ y = \(\frac{1}{x}\) x+y = x + \(\frac{1}{x}\) Let f(x) = x + \(\frac{1}{x}\),x > 0 Differentiating f(x) with respect to x, we get f'(x) = 1 - \(\frac{1}{x^2}\) Also, Differentiating f’(x) with respect to x, we get f''(x) = \(\frac{2}{x^3}\) For minima at x = c, f’(c) = 0 and f’’(c) < 0 ⇒ 1 - \(\frac{1}{x^2}\) = 0 or x = 1 (Since x>0) f’’(1) = 2 > 0 Hence, x = 1 is a point of minima for f(x) and f(1) = 2 is the minimum value of f(x) for x > 0.

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. \(3\frac{1}{3}\)

Answer»

Option : (B)

xy = 1, x > 0, y > 0

⇒ y = \(\frac{1}{x}\)

x+y = x + \(\frac{1}{x}\)

Let f(x) = x + \(\frac{1}{x}\),x > 0

Differentiating f(x) with respect to x, we get

f'(x) = 1 - \(\frac{1}{x^2}\)

Also,

Differentiating f’(x) with respect to x, we get

f''(x) = \(\frac{2}{x^3}\)

For minima at x = c,

f’(c) = 0 and f’’(c) < 0

⇒ 1 - \(\frac{1}{x^2}\) = 0

or x = 1 (Since x>0)

f’’(1) = 2 > 0

Hence,

x = 1 is a point of minima for f(x) and f(1) = 2 is the minimum value of f(x) for x > 0.