Question

Sum of two non zero numbers is 12.what is the minimum sum of their reciprocals?

Answer 1

Let the two numbers are x and y

We have,

x + y = 12 => y = 12 - x

Now,

$\frac{1}{x} + \frac{1}{y} \\ = \frac{x + y}{xy} \\ = \frac{12}{x(12 - x)}$

Thus, we are left with,

$\frac{1}{x} + \frac{1}{y} = \frac{12}{x(12 - x)}$

Now, for the minimum value for 1/x + 1/y, x(12 - x) will be maximum.

So, differentiating x(12 - x) w.r.t. x and then equating the the result to 0 will give us the value of x at which x(12 - x) is maximum.

=> 12 - 2x = 0

=> x = 6

Hence, the maximum value of x(12 - x) =

6(12 - 6) = 36

Therefore, the minimum value of 1/x + 1/y = 12/36 = ⅓

Hope, it'll help you.....