Question

given f(x) = x^2 + 4. for x and y are real positif number thas satisfying f(xy) + f(y − x) = f(y + x). find the maximum value of x + y

Answer 1

@rational

Answer 2

IMO problem again ha

Answer 3

no it is not :D

Answer 4

my work is :
f(xy) = f(y + x) - f(y -x)
(xy)^2 + 4 = (y+x)^2 + 4 -((y-x)^2 + 4)
(xy)^2 + 4 = 2xy + 2xy
(xy)^2 + 4 = 4xy
(xy)^2 - 4xy + 4 = 0
(xy - 2)^2 = 0
xy = 2

Answer 5

what next ?

Answer 6

set x = y ?

Answer 7

looks neat !!!!

Answer 8

yes that will do

Answer 9

maximize x+y
subject to xy=2

Answer 10

so, max of (x+y) = 2 sqrt(2) ?

Answer 11

yes that should work :)

Answer 12

xy=2 ---> y = 2/x

k = x + y
k = x + 2/x
k = (x^2 + 2)/x
k' = (2x(x) - 1(x^2 + 2))/x^2 = 0
2x^2 - x^2 - 2 = 0
x = (1 +- sqrt((-1)^2 - 4(2)(-2))/(2(2))
x = (1 + sqrt(17)/4

k max = ((1 + sqrt(17)/4)^2 + 2)/ ((1 + sqrt(17)/4

Answer 13

hmm... i think there is something wrong with my job

Answer 14

http://www.wolframalpha.com/input/?i=maximize+x%2By%2C+xy%3D2

Answer 15

```
xy=2 ---> y = 2/x

k = x + y
k = x + 2/x
k = (x^2 + 2)/x
k' = (2x(x) - 1(x^2 + 2))/x^2 = 0
2x^2 - x^2 - 2 = 0
```

immediately after this, dont you get `x^2 - 2 = 0` ?

Answer 16

opsss.. yes that's my mistake. that's should be :
x^2 - 2 = 0
x^2 = 2
x = +- sqrt(2)
obvious take x positive, so k max = (x^2 + 2)/x = (2 + 2)/sqrt(2) = 2sqrt(2)
thanks :D

Answer 17

but i dont understand why the shortway is just take x = y ?

Answer 18

thats a geometric argument,

given that area of a rectangle = xy = 2
the largest possible perimeter is achieved for a square shape

Answer 19

oh, ok.. thank you very much :D

Answer 20

np :)

Answer 21

wait, why the wolfram said max is infinite ? --"

Answer 22

is the question should asking us find the minimum value ?

Answer 23

Ahh yes it seems the problem is more involved

Answer 24

you probably know this but i think you made that differentiating task harder than it should have been

Answer 25

\[k=x+\frac{2}{x} \\ k'=1-\frac{2}{x^2} \\ 1-\frac{2}{x^2}=0 \\ 1=\frac{2}{x^2} \\ x^2=2\]