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

1. Astrophysics

Is it $g(x,y) = \frac{ (x^2+y)e^y }{ 2 }$

2. anonymous

no just e^(y/2)

3. anonymous

no y/2 is the exponent of e

4. Astrophysics

$g(x,y)= (x^2+y)e^{y/2}$

5. anonymous

yes

6. Astrophysics

We have to do the second derivative test, you know it?

7. Astrophysics

Lets make it easier on ourselves and expand it $g(x,y) = x^2e^{y/2}+ye^{y/2}$

8. anonymous

ok

9. Astrophysics

So find all the derivatives, find $\huge f_x,f_y, f_{xx}, f_{yy}$

10. Astrophysics

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

But, lets just go over this and you will see, so find the partial derivatives as I've asked you to

12. anonymous

fx= 2xe^(y/2)

13. anonymous

fy=e^(y/2)(x^2 +y /2) + 1

14. anonymous

fxx=2e^(y/2)

15. anonymous

fyy=1/2 e^(y/2) +1/2 e^(y/2) (x^2 +y +2/2)

16. anonymous

@1069592ppl

17. Astrophysics

$\large g_x = 2xe^{y/2} ~~~ \checkmark$ $\large g_{xx} = 2e^{y/2}~~~\checkmark$ That looks good, now let me check your partial respect to y.

18. Astrophysics

$\large g_{y} = \frac{ 1 }{ 2 }e^{y/2} (x^2+y+2)$ this should be function with respect to y

19. Astrophysics

Then $\large g_{yy} = \frac{ 1 }{ 4 }e^{y/2}(x^2+y+4)$

20. anonymous

ok thanks

21. Astrophysics

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

These are long problems..I know haha.

23. Astrophysics

So for gx we are solving for y

24. Astrophysics

$2xe^{y/2}=0$

25. anonymous

xey/2=0

26. anonymous

how do we solve for y

27. Astrophysics

Mhm sec, I am getting - infinity for that

28. Astrophysics

Oh oops, solve for x haha.

29. Astrophysics

That makes more sense, and should be a lot more easier :)

30. anonymous

so x=0

31. Astrophysics

Right, now for gy solve for y

32. Astrophysics

$\large g_y = \frac{ 1 }{ 2 }e^{y/2}(x^2+y+2) = 0$

33. Astrophysics

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

i don't know how to solve it because of e^(y/2)

35. Astrophysics

Get rid of it :P |dw:1438324656542:dw|

36. Astrophysics

Now go ahead and solve for y

37. anonymous

oh both e^(y/2) and (x2+y+2)=0 so y= -2 right?

38. anonymous

so (0,-2) is the critical point

39. Astrophysics

solving for y we get y = -x^2-2

40. Astrophysics

So now if x = 0, y = y = -0^2 -2 = -2 and wow, that seems to be our only critical point

41. Astrophysics

(0,-2)

42. Astrophysics

Sorry I've been lagging a lot so it's hard to work, ok let me see...

43. anonymous

no i really appreciate your work

44. Astrophysics

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

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

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

So find $g_{xy}$ first

48. anonymous

x

49. anonymous

so (2/e) (1/2e)- 0=1/e^2 right?

50. Astrophysics

What's that?

51. anonymous

D

52. Astrophysics

$g_{xy} = xe^{y/2}$

53. Astrophysics

$D = g(0,-2) = g_{xx}(0,-2)g_{yy}(0,-2)-[g_{xy}(0,-2)]^2$

54. anonymous

D= (2/e) (1/2e)- 0=1/e^2

55. Astrophysics

Oh ok, so you plugged it all in :)

56. Astrophysics

So if 1/e^2 < 0 then we have a saddle point

57. anonymous

but it is >0

58. Astrophysics

Right, since 1/e^2 is approximately 0.14 so remember the rules I showed you above, what can we conclude?

59. anonymous

so we see gxx

60. Astrophysics

Good! Now is gxx >0 or <0

61. Astrophysics

By the way, something to note, if D = 0, then the test fails

62. anonymous

oh thanks

63. anonymous

gxx >0 so will get min

64. Astrophysics

Yes a local min :)

65. Zale101

Awesome work @Astrophysics ^_^

66. Astrophysics

Haha, thanks Zale, I just wish it wasn't as laggy

67. Zale101

I can tell. Your icon does not appear on the top thread of this post

68. Astrophysics

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

g(0,-2)= -2/e

70. anonymous

is our local min value

71. anonymous

Thanks a lot !!!

72. Astrophysics

If gxx(0,2) < 0 then it would be local maximum and np :)