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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

Mathematics
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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 ...

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Other answers:

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
but then i tried plugging in the values to the orginal equation and the number seemed wrong so idk what to do
firstly tell me that you want to proceed using concept of maxima minima using partial derivatives of langranges multipliers
is there an easier way
im good with anything
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) ?
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
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?
sry im not exactly sure what you mean
and whered the c come from
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
so 2y=c
3x=c+4
im sry im not sure what to do now
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
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
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
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
yes perhaps then we screwed up with the calculation part can you tell me the answers if you have them with you?
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
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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