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## ranyai12 2 years ago 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

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1. harsimran_hs4

hey have you heard of langranges multipliers or not?

2. ranyai12

yea but i still got it wrong

3. harsimran_hs4

ok so can you show the working ...

4. ranyai12

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

5. ranyai12

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

6. ranyai12

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

7. harsimran_hs4

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

8. ranyai12

is there an easier way

9. ranyai12

im good with anything

10. 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) ?

11. ranyai12

fx=6x-4 fy=4y

12. harsimran_hs4

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

13. ranyai12

im not sure im confused

14. ranyai12

fx=2x and fy=2y

15. ranyai12

so 2xi and 2yj

16. 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?

17. ranyai12

sry im not exactly sure what you mean

18. ranyai12

and whered the c come from

19. 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

20. ranyai12

so 2y=c

21. ranyai12

3x=c+4

22. ranyai12

im sry im not sure what to do now

23. 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

24. ranyai12

so 2 is the max and 0 is the min

25. ranyai12

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

26. 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

27. ranyai12

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

28. ranyai12

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

29. ranyai12

@harsimran_hs4 did you get 1 as well

30. harsimran_hs4

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

31. ranyai12

the answers arent given :(

32. ranyai12

its an online hw thing

33. harsimran_hs4

ok then why did you say it`s wrong?

34. ranyai12

because i plugged t in and it said it was wrong

35. harsimran_hs4

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

36. ranyai12

ok thanks

37. harsimran_hs4

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

38. ranyai12

its not -3

39. 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

40. ranyai12

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

41. harsimran_hs4

then wait...

42. harsimran_hs4

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

43. ranyai12

yea but i can still get partial pints

44. 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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