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

- anonymous

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

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- Astrophysics

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

- anonymous

no just e^(y/2)

- anonymous

no y/2 is the exponent of e

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## More answers

- Astrophysics

\[g(x,y)= (x^2+y)e^{y/2}\]

- anonymous

yes

- Astrophysics

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

- Astrophysics

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

- anonymous

ok

- Astrophysics

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

- 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

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

- anonymous

fx= 2xe^(y/2)

- anonymous

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

- anonymous

fxx=2e^(y/2)

- anonymous

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

- anonymous

@1069592ppl

- 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

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

- Astrophysics

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

- anonymous

ok thanks

- 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

These are long problems..I know haha.

- Astrophysics

So for gx we are solving for y

- Astrophysics

\[2xe^{y/2}=0\]

- anonymous

xey/2=0

- anonymous

how do we solve for y

- Astrophysics

Mhm sec, I am getting - infinity for that

- Astrophysics

Oh oops, solve for x haha.

- Astrophysics

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

- anonymous

so x=0

- Astrophysics

Right, now for gy solve for y

- Astrophysics

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

- 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

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

- Astrophysics

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

- Astrophysics

Now go ahead and solve for y

- anonymous

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

- anonymous

so (0,-2) is the critical point

- Astrophysics

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

- Astrophysics

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

- Astrophysics

(0,-2)

- Astrophysics

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

- anonymous

no i really appreciate your work

- 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

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

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

So find \[g_{xy}\] first

- anonymous

x

- anonymous

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

- Astrophysics

What's that?

- anonymous

D

- Astrophysics

\[g_{xy} = xe^{y/2}\]

- Astrophysics

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

- anonymous

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

- Astrophysics

Oh ok, so you plugged it all in :)

- Astrophysics

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

- anonymous

but it is >0

- Astrophysics

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

- anonymous

so we see gxx

- Astrophysics

Good! Now is gxx >0 or <0

- Astrophysics

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

- anonymous

oh thanks

- anonymous

gxx >0 so will get min

- Astrophysics

Yes a local min :)

- Zale101

Awesome work @Astrophysics ^_^

- Astrophysics

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

- Zale101

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

- 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

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

- anonymous

is our local min value

- anonymous

Thanks a lot !!!

- Astrophysics

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

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