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.