Question

Calculate the Local Extrema

Answer 1

the function is: F(x,y)=1/x+1/y
and the condition is 1/x^2 +1/y^2 = 1

Answer 2

Here's what i did:

Answer 3

I am not sure what to do next, how to determine if d^2F is > or < than 0?

Answer 4

You have two equations :

-1/x^2 -2L/x^3 = 0

-1/y^2 -2L/y^3 = 0

Answer 5

You get

x = -2L
y = -2L

right ?

Answer 6

yes

Answer 7

divide them and get y = x

plug this in the constraint 1/x^2 + 1/y^2 = 1

Answer 8

1/x^2 + 1/x^2 = 1

solve x

Answer 9

where do i divide them?

Answer 10

x = -2L
y = -2L



x/y = ?

Answer 11

well then x/y = 1?
i am not sure why am i doing this?

Answer 12

Your goal is to find x, y that satisfy the 3 equations that you have in lagrange multipliers

Answer 13

Eliminating L gave you the equation y = x

Answer 14

Now eliminate y by replacing y by x in the constraint equation

Answer 15

but i wasn't given that equation..

Answer 16

which equation are you talking about ?

Answer 17

y=x

Answer 18

you've got that equation by eliminating L from the two equations :


-1/x^2 -2L/x^3 = 0

-1/y^2 -2L/y^3 = 0

Answer 19

okay so now i put x as y in the conditional one?

Answer 20

Yes, that gives you an equation with only one variable. Solving that would be easy

Answer 21

so x= + - 1/sqrt2

Answer 22

so from this y also = +- 1/sqrt2
and L=+- 1/(-2sqrt2) ?