Question

Explain Please?

if `x+y=4` for positive integers
x and y , then the smallest possible value of `1/x + 1/y`

Answer 1

1

Explanation:

There are only two possible combinations of positive integer `x` and `y`. These are `2` and `2`, and `3` and `1`. Luckily, this is not too exhaustive to test both of these.

Let `x=2` and `y=2`:

`1/2+1/2=1`

Let `x=3` and `y=1`

`1/3+1/1=4/3`

Therefore, since `1<4/3`, the minimum value is 1

Answer 2

Isn't it 1

Explanation:

`x+y=4`
`=>x=4-y`

Substituting in the equation:
`=1/(4-y)+1/y`

LCM of denominators is `y(4-y)`
`=y/((4-y)y)+(4-y)/(y(4-y))`

Combining the denominators:
`=(y(4-y))/(y(4-y))`

`=1`

Answer 3

Given: `x+y=4, x>0, y>0`

Minimize:

`1/x+1/y`

Substitute `y = 4-x`:

`1/x+1/(4-x)`

Compute the first derivative with respect to x:

`dy/dx = -1/x^2+1/(4-x)^2`

To find local extrema we set the first derivative equal to 0:

`0 = -1/x^2+1/(4-x)^2`

Multiply both sides by `-x^2(4-x)^2`:

`0 = (4-x)^2-x^2`

Expand the square:

`0 = x^2-8x+16-x^2`

`0 = -8x+16`

`x = 2`

Find the corresponding value of y:

`y = 4-2`

`y = 2`

The value is:

`1/2+1/2 = 1`

We could perform the second derivative test but we know that this is a minimum because it is obvious that the maximum occurs when either `x` or `y` approaches 0.