Question

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

Answer 1

hey have you heard of langranges multipliers or not?

Answer 2

yea but i still got it wrong

Answer 3

ok so can you show the working ...

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

is there an easier way

Answer 9

im good with anything

Answer 10

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) ?

Answer 11

fx=6x-4
fy=4y

Answer 12

so that is (6x -4)i + (4y)j

tell me the same for the boundary x^2 + y^2 = 4 also?

Answer 13

im not sure im confused

Answer 14

fx=2x and fy=2y

Answer 15

so 2xi and 2yj

Answer 16

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?

Answer 17

sry im not exactly sure what you mean

Answer 18

and whered the c come from

Answer 19

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

Answer 20

so 2y=c

Answer 21

3x=c+4

Answer 22

im sry im not sure what to do now

Answer 23

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

Answer 24

so 2 is the max and 0 is the min

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

@harsimran_hs4 did you get 1 as well

Answer 30

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

Answer 31

the answers arent given :(

Answer 32

its an online hw thing

Answer 33

ok then why did you say it`s wrong?

Answer 34

because i plugged t in and it said it was wrong

Answer 35

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

Answer 36

ok thanks

Answer 37

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

Answer 38

its not -3

Answer 39

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

Answer 40

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

Answer 41

then wait...

Answer 42

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

Answer 43

yea but i can still get partial pints

Answer 44

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