Can someone please help!!! I keep getting it wrong!!!
Find the extreme values of
f(xy)=3x^2+2y^2−4x−3 on the region described as x^2+y^2 is less than or equal to 4

- anonymous

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- harsimran_hs4

hey have you heard of langranges multipliers or not?

- anonymous

yea but i still got it wrong

- harsimran_hs4

ok so can you show the working ...

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## More answers

- anonymous

yea this is what i did:
f_x=6x-4<=4
6x<=8
x<=4/3
f_y=4y<=4 f_y<=1
fxx=6
fyy=4

- anonymous

and then i got stuck so i assumed the max was 6 and min was 1

- anonymous

but then i tried plugging in the values to the orginal equation and the number seemed wrong so idk what to do

- harsimran_hs4

firstly tell me that you want to proceed using concept of maxima minima using partial derivatives of langranges multipliers

- anonymous

is there an easier way

- anonymous

im good with anything

- harsimran_hs4

well perhaps i don`t think so because it`s gone in 2 dimension
so lets do it by langranges multipliers
1. tell me what is gradient of f(x,y) ?

- anonymous

fx=6x-4
fy=4y

- harsimran_hs4

so that is (6x -4)i + (4y)j
tell me the same for the boundary x^2 + y^2 = 4 also?

- anonymous

im not sure im confused

- anonymous

fx=2x and fy=2y

- anonymous

so 2xi and 2yj

- harsimran_hs4

yes that is fine
(2x)i + (2y)j
so for extreme values to occur the gradient of the function and that of the boundary must be proportional
so
(6x -4)i + (4y)j = c((2x)i + (2y)j)
now compare the i and j components and find the value of x and y in terms of c and tell me?

- anonymous

sry im not exactly sure what you mean

- anonymous

and whered the c come from

- harsimran_hs4

6x - 4 = 2cx
and
4y = 2cy
c is the proportionality constant
better say the extreme values occurs when the gradient of two is antiparallel or proportional

- anonymous

so 2y=c

- anonymous

3x=c+4

- anonymous

im sry im not sure what to do now

- harsimran_hs4

x = 4/(6 - 2c)
and c = 2 from second equation
so x = 2
not the boundary x^2 + y^2 = 4
putting x = 2
and finding out value of y which come out to be 0
so (2, 0) is the point of extreme value

- anonymous

so 2 is the max and 0 is the min

- anonymous

or do we plus in 2 into the equatio and thats hte max and 0 is the min

- harsimran_hs4

no the point (2, 0) will result in the extreme value of the function
put x = 2 and y = 0 to yield the extreme value

- anonymous

but wouldnt that result in just one value and not max and min

- anonymous

yr supposed to find absolute max and min and i got the answer ot be 1 which is wrong

- anonymous

@harsimran_hs4 did you get 1 as well

- harsimran_hs4

yes perhaps then we screwed up with the calculation part
can you tell me the answers if you have them with you?

- anonymous

the answers arent given :(

- anonymous

its an online hw thing

- harsimran_hs4

ok then why did you say it`s wrong?

- anonymous

because i plugged t in and it said it was wrong

- harsimran_hs4

ok
let me figure out then and i`ll get back to you

- anonymous

ok thanks

- harsimran_hs4

can you plug in -3 and check if thats minimum value?

- anonymous

its not -3

- harsimran_hs4

do you get -ve points if you put in a wrong answer
if no then check out if 9 is maximum or not

- anonymous

i only have one more try left to find out what the max and min is

- harsimran_hs4

then wait...

- harsimran_hs4

you need to plug in both the values simultaneously or not?

- anonymous

yea but i can still get partial pints

- harsimran_hs4

http://www.math.uri.edu/~bkaskosz/flashmo/graph3d/
check this
this will give some insight to the problem
but do link someone else also to this problem and figure out...
i`ll try to get to some sure answer and then reply

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