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anonymous

  • one year ago

Find the local minimum, maximum, and saddle points of the function g(x,y) = (x2 + y) e^y/2 .

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  1. Astrophysics
    • one year ago
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    Is it \[g(x,y) = \frac{ (x^2+y)e^y }{ 2 }\]

  2. anonymous
    • one year ago
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    no just e^(y/2)

  3. anonymous
    • one year ago
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    no y/2 is the exponent of e

  4. Astrophysics
    • one year ago
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    \[g(x,y)= (x^2+y)e^{y/2}\]

  5. anonymous
    • one year ago
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    yes

  6. Astrophysics
    • one year ago
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    We have to do the second derivative test, you know it?

  7. Astrophysics
    • one year ago
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    Lets make it easier on ourselves and expand it \[g(x,y) = x^2e^{y/2}+ye^{y/2}\]

  8. anonymous
    • one year ago
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    ok

  9. Astrophysics
    • one year ago
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    So find all the derivatives, find \[\huge f_x,f_y, f_{xx}, f_{yy}\]

  10. Astrophysics
    • one year ago
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    Essentially the second derivative rule says, let \[D = D(a,b) = f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2\] and if D>0 and fxx(a,b) > 0, then f(a,b) is a local min, if D>0, fxx(a,b)<0 then f(a,b) is a local maximum, and if D<0 then f(a,b) then it is a saddle point as it is neither max/ min.

  11. Astrophysics
    • one year ago
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    But, lets just go over this and you will see, so find the partial derivatives as I've asked you to

  12. anonymous
    • one year ago
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    fx= 2xe^(y/2)

  13. anonymous
    • one year ago
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    fy=e^(y/2)(x^2 +y /2) + 1

  14. anonymous
    • one year ago
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    fxx=2e^(y/2)

  15. anonymous
    • one year ago
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    fyy=1/2 e^(y/2) +1/2 e^(y/2) (x^2 +y +2/2)

  16. anonymous
    • one year ago
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    @1069592ppl

  17. Astrophysics
    • one year ago
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    \[\large g_x = 2xe^{y/2} ~~~ \checkmark\] \[\large g_{xx} = 2e^{y/2}~~~\checkmark\] That looks good, now let me check your partial respect to y.

  18. Astrophysics
    • one year ago
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    \[\large g_{y} = \frac{ 1 }{ 2 }e^{y/2} (x^2+y+2)\] this should be function with respect to y

  19. Astrophysics
    • one year ago
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    Then \[\large g_{yy} = \frac{ 1 }{ 4 }e^{y/2}(x^2+y+4)\]

  20. anonymous
    • one year ago
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    ok thanks

  21. Astrophysics
    • one year ago
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    Now since we have all the partial derivatives, now we need to see where \[\large g_x = 0~~~\text{and}~~~g_y = 0\] So lets set it up as such \[\large g_x=2xe^{y/2} =0\] and \[g_y = \frac{ 1 }{ 2 }e^{y/2}(x^2+y+2) = 0\] we set them equal to 0, note we are looking for the conditions now which satisfy both of the equations.

  22. Astrophysics
    • one year ago
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    These are long problems..I know haha.

  23. Astrophysics
    • one year ago
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    So for gx we are solving for y

  24. Astrophysics
    • one year ago
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    \[2xe^{y/2}=0\]

  25. anonymous
    • one year ago
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    xey/2=0

  26. anonymous
    • one year ago
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    how do we solve for y

  27. Astrophysics
    • one year ago
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    Mhm sec, I am getting - infinity for that

  28. Astrophysics
    • one year ago
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    Oh oops, solve for x haha.

  29. Astrophysics
    • one year ago
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    That makes more sense, and should be a lot more easier :)

  30. anonymous
    • one year ago
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    so x=0

  31. Astrophysics
    • one year ago
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    Right, now for gy solve for y

  32. Astrophysics
    • one year ago
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    \[\large g_y = \frac{ 1 }{ 2 }e^{y/2}(x^2+y+2) = 0\]

  33. Astrophysics
    • one year ago
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    So the reason we're doing this is actually to find the critical points, this is the most tedious part, but after it will get better.

  34. anonymous
    • one year ago
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    i don't know how to solve it because of e^(y/2)

  35. Astrophysics
    • one year ago
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    Get rid of it :P |dw:1438324656542:dw|

  36. Astrophysics
    • one year ago
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    Now go ahead and solve for y

  37. anonymous
    • one year ago
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    oh both e^(y/2) and (x2+y+2)=0 so y= -2 right?

  38. anonymous
    • one year ago
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    so (0,-2) is the critical point

  39. Astrophysics
    • one year ago
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    solving for y we get y = -x^2-2

  40. Astrophysics
    • one year ago
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    So now if x = 0, y = y = -0^2 -2 = -2 and wow, that seems to be our only critical point

  41. Astrophysics
    • one year ago
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    (0,-2)

  42. Astrophysics
    • one year ago
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    Sorry I've been lagging a lot so it's hard to work, ok let me see...

  43. anonymous
    • one year ago
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    no i really appreciate your work

  44. Astrophysics
    • one year ago
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    Even if we plug in for y in the first equation it wouldn't really matter, we would still only get 1 critical value, mhm.

  45. Astrophysics
    • one year ago
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    So what I showed you first, \[D = D(a,b) = f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2\] this is what we use to test to see our points.

  46. Astrophysics
    • one year ago
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    Since we only have 1 critical value, that should be easy, kind of weird how we only got one, but I don't think we did anything wrong. So lets test @(0,-2)

  47. Astrophysics
    • one year ago
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    So find \[g_{xy}\] first

  48. anonymous
    • one year ago
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    x

  49. anonymous
    • one year ago
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    so (2/e) (1/2e)- 0=1/e^2 right?

  50. Astrophysics
    • one year ago
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    What's that?

  51. anonymous
    • one year ago
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    D

  52. Astrophysics
    • one year ago
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    \[g_{xy} = xe^{y/2}\]

  53. Astrophysics
    • one year ago
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    \[D = g(0,-2) = g_{xx}(0,-2)g_{yy}(0,-2)-[g_{xy}(0,-2)]^2\]

  54. anonymous
    • one year ago
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    D= (2/e) (1/2e)- 0=1/e^2

  55. Astrophysics
    • one year ago
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    Oh ok, so you plugged it all in :)

  56. Astrophysics
    • one year ago
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    So if 1/e^2 < 0 then we have a saddle point

  57. anonymous
    • one year ago
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    but it is >0

  58. Astrophysics
    • one year ago
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    Right, since 1/e^2 is approximately 0.14 so remember the rules I showed you above, what can we conclude?

  59. anonymous
    • one year ago
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    so we see gxx

  60. Astrophysics
    • one year ago
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    Good! Now is gxx >0 or <0

  61. Astrophysics
    • one year ago
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    By the way, something to note, if D = 0, then the test fails

  62. anonymous
    • one year ago
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    oh thanks

  63. anonymous
    • one year ago
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    gxx >0 so will get min

  64. Astrophysics
    • one year ago
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    Yes a local min :)

  65. Zale101
    • one year ago
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    Awesome work @Astrophysics ^_^

  66. Astrophysics
    • one year ago
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    Haha, thanks Zale, I just wish it wasn't as laggy

  67. Zale101
    • one year ago
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    I can tell. Your icon does not appear on the top thread of this post

  68. Astrophysics
    • one year ago
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    XD, yup I've already lost connection twice during this question. So if you notice @Hano these problems are quite long, but it's good experience! So I hope you understood the process and why we did what :)

  69. anonymous
    • one year ago
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    g(0,-2)= -2/e

  70. anonymous
    • one year ago
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    is our local min value

  71. anonymous
    • one year ago
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    Thanks a lot !!!

  72. Astrophysics
    • one year ago
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    If gxx(0,2) < 0 then it would be local maximum and np :)

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