Question

evaluate the function for relative extrema. evaluate the function for relative extrema. @Mathematics

Answer 1

\[\sqrt{x^2+y^2}\]

Answer 2

Is it just me or is the second partial test inconclusive

Answer 3

so you want to find the relative extrema?

Answer 4

yeah i did all the second partials and stuff but D=0 when i do it

Answer 5

they say there is a relative min at (0,0,0)

Answer 6

ok one sec

Answer 7

that's correct

Answer 8

that's actually a global min

Answer 9

so does d=0 or something else

Answer 10

but you have the right idea

Answer 11

not sure what you're referring to with d

Answer 12

not sure what it means... lol my teacher never said the name of it

Fxx*Fyy-Fxy

Answer 13

\[F_ {xx}*F_ {yy}-[F_ {xy}]^2=d\]

Answer 14

ah found what you mean

Answer 15

yeah my teacher was calling it distance for a while but then we got confused when he was using the distance formula lol

Answer 16

D = fxx(a,b)*fyy(a,b) - {fxy(a,b)}², then

1) D > 0 and fxx > 0 => min
2) D > 0 and fxx < 0 => max
3) D < 0 => saddle point
4) D = 0 - indeterminate

Answer 17

found here: http://www.physicsforums.com/showthread.php?t=195846

Answer 18

i think he meant discriminant...but not sure tbh

Answer 19

alright so when you did the second partials didn't you get 0*0-0^2=0

Answer 20

let me differentiate real quick

Answer 21

no matter what i always get x and y variables on top which will always = 0 but when you also use fxx,fxy,fyy(0,0,z) you get indeterminates

Answer 22

yeah I'm getting division by zero errors

Answer 23

well when you put it into form of a surface you get

0=x^2+y^2-z^2 which is going to be a point

Answer 24

let me double check my work again, something might've gone wrong

Answer 25

which doesn't work with a hyperboloid... so i'm confusedl ol

Answer 26

if i trace it maybe...

Answer 27

i think I know what's wrong

Answer 28

?

Answer 29

the theorem shown above only works with continuous partial second derivatives at (x,y)

Answer 30

so how suppose i solve this?

Answer 31

using the range?

Answer 32

you're probably right in saying you have to work with 0=x^2+y^2-z^2

Answer 33

just not sure how yet

Answer 34

knowing that the range must be >=0 since it's square so the lowest it can be is 0?

Answer 35

that's a good way to put it

Answer 36

the thing is... that isn't a quadric surface ?

Answer 37

the closest it is to is a hyperboloid of one sheet however that is not a hyperboloid of one sheet

Answer 38

it looks sort of like a paraboloid

Answer 39

paraboloid would be z = x^2+y^2 right?

Answer 40

yeah it's a hyperboloid of one sheet

Answer 41

not according to wolfram alpha?

Answer 42

http://www.wolframalpha.com/input/?i=f%28x%2Cy%29%3Dsqrt%28x%5E2%2By%5E2%29

Answer 43

I'm looking at this

http://en.wikipedia.org/wiki/Hyperboloid

Answer 44

it's the first line