anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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harsimran_hs4
  • harsimran_hs4
hey have you heard of langranges multipliers or not?
anonymous
  • anonymous
yea but i still got it wrong
harsimran_hs4
  • harsimran_hs4
ok so can you show the working ...

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anonymous
  • 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
  • anonymous
and then i got stuck so i assumed the max was 6 and min was 1
anonymous
  • 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
  • harsimran_hs4
firstly tell me that you want to proceed using concept of maxima minima using partial derivatives of langranges multipliers
anonymous
  • anonymous
is there an easier way
anonymous
  • anonymous
im good with anything
harsimran_hs4
  • 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
  • anonymous
fx=6x-4 fy=4y
harsimran_hs4
  • harsimran_hs4
so that is (6x -4)i + (4y)j tell me the same for the boundary x^2 + y^2 = 4 also?
anonymous
  • anonymous
im not sure im confused
anonymous
  • anonymous
fx=2x and fy=2y
anonymous
  • anonymous
so 2xi and 2yj
harsimran_hs4
  • 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
  • anonymous
sry im not exactly sure what you mean
anonymous
  • anonymous
and whered the c come from
harsimran_hs4
  • 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
  • anonymous
so 2y=c
anonymous
  • anonymous
3x=c+4
anonymous
  • anonymous
im sry im not sure what to do now
harsimran_hs4
  • 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
  • anonymous
so 2 is the max and 0 is the min
anonymous
  • anonymous
or do we plus in 2 into the equatio and thats hte max and 0 is the min
harsimran_hs4
  • 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
  • anonymous
but wouldnt that result in just one value and not max and min
anonymous
  • anonymous
yr supposed to find absolute max and min and i got the answer ot be 1 which is wrong
anonymous
  • anonymous
@harsimran_hs4 did you get 1 as well
harsimran_hs4
  • 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
  • anonymous
the answers arent given :(
anonymous
  • anonymous
its an online hw thing
harsimran_hs4
  • harsimran_hs4
ok then why did you say it`s wrong?
anonymous
  • anonymous
because i plugged t in and it said it was wrong
harsimran_hs4
  • harsimran_hs4
ok let me figure out then and i`ll get back to you
anonymous
  • anonymous
ok thanks
harsimran_hs4
  • harsimran_hs4
can you plug in -3 and check if thats minimum value?
anonymous
  • anonymous
its not -3
harsimran_hs4
  • 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
  • anonymous
i only have one more try left to find out what the max and min is
harsimran_hs4
  • harsimran_hs4
then wait...
harsimran_hs4
  • harsimran_hs4
you need to plug in both the values simultaneously or not?
anonymous
  • anonymous
yea but i can still get partial pints
harsimran_hs4
  • 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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