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

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

Mathematics
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Is it \[g(x,y) = \frac{ (x^2+y)e^y }{ 2 }\]
no just e^(y/2)
no y/2 is the exponent of e

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

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