anonymous
  • anonymous
Find the local minimum, maximum, and saddle points of the function g(x,y) = (x2 + y) e^y/2 .
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Astrophysics
  • Astrophysics
Is it \[g(x,y) = \frac{ (x^2+y)e^y }{ 2 }\]
anonymous
  • anonymous
no just e^(y/2)
anonymous
  • anonymous
no y/2 is the exponent of e

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Astrophysics
  • Astrophysics
\[g(x,y)= (x^2+y)e^{y/2}\]
anonymous
  • anonymous
yes
Astrophysics
  • Astrophysics
We have to do the second derivative test, you know it?
Astrophysics
  • Astrophysics
Lets make it easier on ourselves and expand it \[g(x,y) = x^2e^{y/2}+ye^{y/2}\]
anonymous
  • anonymous
ok
Astrophysics
  • Astrophysics
So find all the derivatives, find \[\huge f_x,f_y, f_{xx}, f_{yy}\]
Astrophysics
  • 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.
Astrophysics
  • Astrophysics
But, lets just go over this and you will see, so find the partial derivatives as I've asked you to
anonymous
  • anonymous
fx= 2xe^(y/2)
anonymous
  • anonymous
fy=e^(y/2)(x^2 +y /2) + 1
anonymous
  • anonymous
fxx=2e^(y/2)
anonymous
  • anonymous
fyy=1/2 e^(y/2) +1/2 e^(y/2) (x^2 +y +2/2)
anonymous
  • anonymous
@1069592ppl
Astrophysics
  • 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.
Astrophysics
  • Astrophysics
\[\large g_{y} = \frac{ 1 }{ 2 }e^{y/2} (x^2+y+2)\] this should be function with respect to y
Astrophysics
  • Astrophysics
Then \[\large g_{yy} = \frac{ 1 }{ 4 }e^{y/2}(x^2+y+4)\]
anonymous
  • anonymous
ok thanks
Astrophysics
  • 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.
Astrophysics
  • Astrophysics
These are long problems..I know haha.
Astrophysics
  • Astrophysics
So for gx we are solving for y
Astrophysics
  • Astrophysics
\[2xe^{y/2}=0\]
anonymous
  • anonymous
xey/2=0
anonymous
  • anonymous
how do we solve for y
Astrophysics
  • Astrophysics
Mhm sec, I am getting - infinity for that
Astrophysics
  • Astrophysics
Oh oops, solve for x haha.
Astrophysics
  • Astrophysics
That makes more sense, and should be a lot more easier :)
anonymous
  • anonymous
so x=0
Astrophysics
  • Astrophysics
Right, now for gy solve for y
Astrophysics
  • Astrophysics
\[\large g_y = \frac{ 1 }{ 2 }e^{y/2}(x^2+y+2) = 0\]
Astrophysics
  • 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.
anonymous
  • anonymous
i don't know how to solve it because of e^(y/2)
Astrophysics
  • Astrophysics
Get rid of it :P |dw:1438324656542:dw|
Astrophysics
  • Astrophysics
Now go ahead and solve for y
anonymous
  • anonymous
oh both e^(y/2) and (x2+y+2)=0 so y= -2 right?
anonymous
  • anonymous
so (0,-2) is the critical point
Astrophysics
  • Astrophysics
solving for y we get y = -x^2-2
Astrophysics
  • Astrophysics
So now if x = 0, y = y = -0^2 -2 = -2 and wow, that seems to be our only critical point
Astrophysics
  • Astrophysics
(0,-2)
Astrophysics
  • Astrophysics
Sorry I've been lagging a lot so it's hard to work, ok let me see...
anonymous
  • anonymous
no i really appreciate your work
Astrophysics
  • 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.
Astrophysics
  • 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.
Astrophysics
  • 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)
Astrophysics
  • Astrophysics
So find \[g_{xy}\] first
anonymous
  • anonymous
x
anonymous
  • anonymous
so (2/e) (1/2e)- 0=1/e^2 right?
Astrophysics
  • Astrophysics
What's that?
anonymous
  • anonymous
D
Astrophysics
  • Astrophysics
\[g_{xy} = xe^{y/2}\]
Astrophysics
  • Astrophysics
\[D = g(0,-2) = g_{xx}(0,-2)g_{yy}(0,-2)-[g_{xy}(0,-2)]^2\]
anonymous
  • anonymous
D= (2/e) (1/2e)- 0=1/e^2
Astrophysics
  • Astrophysics
Oh ok, so you plugged it all in :)
Astrophysics
  • Astrophysics
So if 1/e^2 < 0 then we have a saddle point
anonymous
  • anonymous
but it is >0
Astrophysics
  • Astrophysics
Right, since 1/e^2 is approximately 0.14 so remember the rules I showed you above, what can we conclude?
anonymous
  • anonymous
so we see gxx
Astrophysics
  • Astrophysics
Good! Now is gxx >0 or <0
Astrophysics
  • Astrophysics
By the way, something to note, if D = 0, then the test fails
anonymous
  • anonymous
oh thanks
anonymous
  • anonymous
gxx >0 so will get min
Astrophysics
  • Astrophysics
Yes a local min :)
Zale101
  • Zale101
Awesome work @Astrophysics ^_^
Astrophysics
  • Astrophysics
Haha, thanks Zale, I just wish it wasn't as laggy
Zale101
  • Zale101
I can tell. Your icon does not appear on the top thread of this post
Astrophysics
  • 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 :)
anonymous
  • anonymous
g(0,-2)= -2/e
anonymous
  • anonymous
is our local min value
anonymous
  • anonymous
Thanks a lot !!!
Astrophysics
  • Astrophysics
If gxx(0,2) < 0 then it would be local maximum and np :)

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