Question

find the max/min/saddle points

f(x,y) = x/y + 1/x + 1/y

Answer 1

same thing? how do i find critical points?

Answer 2

did you find fx and fy?

Answer 3

yeah

Answer 4

fx=y - 1/x^2 fy=x - 1/y^2

Answer 5

but when i set them equal, it doesn't give me anything

Answer 6

i don't like your fy

Answer 7

or your fx

Answer 8

wait wait sorry the original equation is xy + 1/x + 1/y not x/y

Answer 9

ok

Answer 10

so if i set them equal, x=y and then x and y can be -1 and 0?

Answer 11

\[y-\frac{1}{x^2}=0\]
\[y=\frac{1}{x^2}\]
and so

\[x-\frac{1}{y^2}=0\]

\[x-\frac{1}{1/x^4}=0\]
\[x-x^4=0\]
\[x(1-x^3)=0\]
so x=1,0

but x cant be zero since 0 is not in the domain of the original function.

Answer 12

and if x=1 then y=1

Answer 13

so the only critical point is (1,1)?

Answer 14

correct

Answer 15

and its a saddle?

Answer 16

looks like a min

Answer 17

but i got that it was <0?

Answer 18

what did you get for D?

Answer 19

-1

Answer 20

D=fxx(1,1)fyy(1,1)-[fxy(1,1)]^2

Answer 21

ohh shoot i did it wrong

Answer 22

I'm using my calculator and i get 3

Answer 23

now i get 1

Answer 24

fxx = 1, fyy=1 and fxy=0

Answer 25

fxx=2/x^3
fyy=2/y^3
fxy=1

Answer 26

ahhhh i didn't do chain rule, wow i'm on a role this morning

Answer 27

okay got it thank you!

Answer 28

yw