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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 .
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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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7Is it \[g(x,y) = \frac{ (x^2+y)e^y }{ 2 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no y/2 is the exponent of e

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7\[g(x,y)= (x^2+y)e^{y/2}\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7We have to do the second derivative test, you know it?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7Lets make it easier on ourselves and expand it \[g(x,y) = x^2e^{y/2}+ye^{y/2}\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7So find all the derivatives, find \[\huge f_x,f_y, f_{xx}, f_{yy}\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7Essentially 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
 one year ago
Best ResponseYou've already chosen the best response.7But, lets just go over this and you will see, so find the partial derivatives as I've asked you to

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0fy=e^(y/2)(x^2 +y /2) + 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0fyy=1/2 e^(y/2) +1/2 e^(y/2) (x^2 +y +2/2)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7\[\large g_x = 2xe^{y/2} ~~~ \checkmark\] \[\large g_{xx} = 2e^{y/2}~~~\checkmark\] That looks good, now let me check your partial respect to y.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7\[\large g_{y} = \frac{ 1 }{ 2 }e^{y/2} (x^2+y+2)\] this should be function with respect to y

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7Then \[\large g_{yy} = \frac{ 1 }{ 4 }e^{y/2}(x^2+y+4)\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7Now 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
 one year ago
Best ResponseYou've already chosen the best response.7These are long problems..I know haha.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7So for gx we are solving for y

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how do we solve for y

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7Mhm sec, I am getting  infinity for that

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7Oh oops, solve for x haha.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7That makes more sense, and should be a lot more easier :)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7Right, now for gy solve for y

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7\[\large g_y = \frac{ 1 }{ 2 }e^{y/2}(x^2+y+2) = 0\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7So 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
 one year ago
Best ResponseYou've already chosen the best response.0i don't know how to solve it because of e^(y/2)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7Get rid of it :P dw:1438324656542:dw

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7Now go ahead and solve for y

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh both e^(y/2) and (x2+y+2)=0 so y= 2 right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so (0,2) is the critical point

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7solving for y we get y = x^22

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7So now if x = 0, y = y = 0^2 2 = 2 and wow, that seems to be our only critical point

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7Sorry I've been lagging a lot so it's hard to work, ok let me see...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no i really appreciate your work

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7Even 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
 one year ago
Best ResponseYou've already chosen the best response.7So 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
 one year ago
Best ResponseYou've already chosen the best response.7Since 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
 one year ago
Best ResponseYou've already chosen the best response.7So find \[g_{xy}\] first

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so (2/e) (1/2e) 0=1/e^2 right?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7\[g_{xy} = xe^{y/2}\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7\[D = g(0,2) = g_{xx}(0,2)g_{yy}(0,2)[g_{xy}(0,2)]^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0D= (2/e) (1/2e) 0=1/e^2

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7Oh ok, so you plugged it all in :)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7So if 1/e^2 < 0 then we have a saddle point

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7Right, since 1/e^2 is approximately 0.14 so remember the rules I showed you above, what can we conclude?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7Good! Now is gxx >0 or <0

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7By the way, something to note, if D = 0, then the test fails

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0gxx >0 so will get min

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7Yes a local min :)

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0Awesome work @Astrophysics ^_^

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7Haha, thanks Zale, I just wish it wasn't as laggy

Zale101
 one year ago
Best ResponseYou've already chosen the best response.0I can tell. Your icon does not appear on the top thread of this post

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7XD, 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
 one year ago
Best ResponseYou've already chosen the best response.0is our local min value

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.7If gxx(0,2) < 0 then it would be local maximum and np :)
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