x and y are positive rational numbers such that x+y=5. What is the smallest possible... please read below.?

Question

x and y are positive rational numbers such that x+y=5. What is the smallest possible value of the expression 1/x + 1/y?

Answer 1

x = 5 - y
Define
f(y) = (1/x) + (1/y) = (1 / (5-y) ) + 1/y
f(y) = 5 / (5y - y^2)

f ' (y) =
25(2y - 5)
---------
y^2 (5 - y)^2

Critical points are y = 5/2, 0, 5. Check the sign of f ' (y) in between these values.

f ' (1) = 25(2 - 5)/(+) < 0.
Thus f is decreasing for all 0 < y < 5/2

f ' ( 3) = 25 (6 - 5) / (+) > 0
Thus f is increasing for all 5/2 < y < 5

f ' (10) = 25( 20 - 5) / (+) > 0
Thus, f is increasing for all y > 5

Therefore, f has a minumum at y = 5/2
x = 5 - (5/2)
x = 5/2

(1/x) + (1/y) = (2/5) + 2/5
Answer: 4/5

Answer 2

1/x + 1/y = (x + y) / xy . (x + y) is constant and xy varies.

So we are asking for the largest value of xy and this occurs

when the variables are equal in a constant sum.

So, x = y = 5/2 and 1/x + 1/y = 4/5.

Answer 3

x + y = 5
y = 5-x

let a = 1/x + 1/y = 1/x + 1/(5-x)

a= (5-x)/x(5-x) + x/x(5-x)
a = 5/[x(5-x)] = 5/(5x - x^2)

5x - x^2 = 5/a

Differentiate both sides:
(5 -2x) dx = (-5/a^2) da

We need da/dx = 0 to find critical points of a as a function of x.

da/dx = (5-2x)/(-5/a^2)
we seek:
0 = (5-2x)/(-5/a^2) = -[(5-2x)a^2]/5

either a^2 = 0 or 5-2x = 0

a^2=0 is impossible since neither x nor y can grow without bound.
therefore we are left with 5-2x = 0
giving us x = 5/2 = y

a = 2/5 + 2/5 = 4/5 = 0.8

Answer 4

xy = sixteen => (x, y) = (a million, sixteen), (2, 8), ( 4, 4) whence their sum 4 + 4 = 8 is least 2 numbers are 4, 4. alternately sqrt(manufactured from 2 numbers) supplies the respond for his or her sum being least. In case, sqrt isn't necessary, nearest to it are the respond.