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ranyai12 Group Title

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

  • one year ago
  • one year ago

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  1. harsimran_hs4 Group Title
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    hey have you heard of langranges multipliers or not?

    • one year ago
  2. ranyai12 Group Title
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    yea but i still got it wrong

    • one year ago
  3. harsimran_hs4 Group Title
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    ok so can you show the working ...

    • one year ago
  4. ranyai12 Group Title
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    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

    • one year ago
  5. ranyai12 Group Title
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    and then i got stuck so i assumed the max was 6 and min was 1

    • one year ago
  6. ranyai12 Group Title
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    but then i tried plugging in the values to the orginal equation and the number seemed wrong so idk what to do

    • one year ago
  7. harsimran_hs4 Group Title
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    firstly tell me that you want to proceed using concept of maxima minima using partial derivatives of langranges multipliers

    • one year ago
  8. ranyai12 Group Title
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    is there an easier way

    • one year ago
  9. ranyai12 Group Title
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    im good with anything

    • one year ago
  10. harsimran_hs4 Group Title
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    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) ?

    • one year ago
  11. ranyai12 Group Title
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    fx=6x-4 fy=4y

    • one year ago
  12. harsimran_hs4 Group Title
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    so that is (6x -4)i + (4y)j tell me the same for the boundary x^2 + y^2 = 4 also?

    • one year ago
  13. ranyai12 Group Title
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    im not sure im confused

    • one year ago
  14. ranyai12 Group Title
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    fx=2x and fy=2y

    • one year ago
  15. ranyai12 Group Title
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    so 2xi and 2yj

    • one year ago
  16. harsimran_hs4 Group Title
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    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?

    • one year ago
  17. ranyai12 Group Title
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    sry im not exactly sure what you mean

    • one year ago
  18. ranyai12 Group Title
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    and whered the c come from

    • one year ago
  19. harsimran_hs4 Group Title
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    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

    • one year ago
  20. ranyai12 Group Title
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    so 2y=c

    • one year ago
  21. ranyai12 Group Title
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    3x=c+4

    • one year ago
  22. ranyai12 Group Title
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    im sry im not sure what to do now

    • one year ago
  23. harsimran_hs4 Group Title
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    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

    • one year ago
  24. ranyai12 Group Title
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    so 2 is the max and 0 is the min

    • one year ago
  25. ranyai12 Group Title
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    or do we plus in 2 into the equatio and thats hte max and 0 is the min

    • one year ago
  26. harsimran_hs4 Group Title
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    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

    • one year ago
  27. ranyai12 Group Title
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    but wouldnt that result in just one value and not max and min

    • one year ago
  28. ranyai12 Group Title
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    yr supposed to find absolute max and min and i got the answer ot be 1 which is wrong

    • one year ago
  29. ranyai12 Group Title
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    @harsimran_hs4 did you get 1 as well

    • one year ago
  30. harsimran_hs4 Group Title
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    yes perhaps then we screwed up with the calculation part can you tell me the answers if you have them with you?

    • one year ago
  31. ranyai12 Group Title
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    the answers arent given :(

    • one year ago
  32. ranyai12 Group Title
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    its an online hw thing

    • one year ago
  33. harsimran_hs4 Group Title
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    ok then why did you say it`s wrong?

    • one year ago
  34. ranyai12 Group Title
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    because i plugged t in and it said it was wrong

    • one year ago
  35. harsimran_hs4 Group Title
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    ok let me figure out then and i`ll get back to you

    • one year ago
  36. ranyai12 Group Title
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    ok thanks

    • one year ago
  37. harsimran_hs4 Group Title
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    can you plug in -3 and check if thats minimum value?

    • one year ago
  38. ranyai12 Group Title
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    its not -3

    • one year ago
  39. harsimran_hs4 Group Title
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    do you get -ve points if you put in a wrong answer if no then check out if 9 is maximum or not

    • one year ago
  40. ranyai12 Group Title
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    i only have one more try left to find out what the max and min is

    • one year ago
  41. harsimran_hs4 Group Title
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    then wait...

    • one year ago
  42. harsimran_hs4 Group Title
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    you need to plug in both the values simultaneously or not?

    • one year ago
  43. ranyai12 Group Title
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    yea but i can still get partial pints

    • one year ago
  44. harsimran_hs4 Group Title
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    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

    • one year ago
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