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hey have you heard of langranges multipliers or not?

yea but i still got it wrong

ok so can you show the working ...

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

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

is there an easier way

im good with anything

fx=6x-4
fy=4y

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

im not sure im confused

fx=2x and fy=2y

so 2xi and 2yj

sry im not exactly sure what you mean

and whered the c come from

so 2y=c

3x=c+4

im sry im not sure what to do now

so 2 is the max and 0 is the min

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

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

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

@harsimran_hs4 did you get 1 as well

the answers arent given :(

its an online hw thing

ok then why did you say it`s wrong?

because i plugged t in and it said it was wrong

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

ok thanks

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

its not -3

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

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

then wait...

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

yea but i can still get partial pints