Explain Please?

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

3 Answers
May 13, 2018

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

May 13, 2018

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

May 13, 2018

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.