The sum of two numbers is 6. The minimum value of the sum of their reciprocals is

Options

A. `6/5`
B. `3/4`
C. `2/3`
D. `1/2`

Answer

C

Explanation

Let two numbers be x and y, then
`x+y=6`……….i
Let `z=1/x+1/y` ……………….ii
`impliesz=1/x+1/(6-x)` [ From Eq. i]
`implies(dz)/(dx)=-1/(x^(2))+1/((6-x)^(2))`
Put `(dz)/(dx)=0implies-1/(x^(2))+1/((6-x)^(2))=0`
`implies-(6-x)^(2)+x^(2)=0`
`implies 12x-36=0impliesx=3`
Now, `(d^(2)z)/(dx^(2))=2/(x^(3))+2/((6-x)^(3))`
At `x=3, (d^(2)z)/(dx^(2))gt0,` minimum.
Hence, minimum value at `x=3` is
`z=1/3+1/3=2/3`