Question

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

Answer 1

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

Answer 2

no just e^(y/2)

Answer 3

no y/2 is the exponent of e

Answer 4

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

Answer 5

yes

Answer 6

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

Answer 7

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

Answer 8

ok

Answer 9

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

Answer 10

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.

Answer 11

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

Answer 12

fx= 2xe^(y/2)

Answer 13

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

Answer 14

fxx=2e^(y/2)

Answer 15

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

Answer 16

@1069592ppl

Answer 17

\[\large g_x = 2xe^{y/2} ~~~ \checkmark\]
\[\large g_{xx} = 2e^{y/2}~~~\checkmark\]

That looks good, now let me check your partial respect to y.

Answer 18

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

Answer 19

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

Answer 20

ok thanks

Answer 21

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.

Answer 22

These are long problems..I know haha.

Answer 23

So for gx we are solving for y

Answer 24

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

Answer 25

xey/2=0

Answer 26

how do we solve for y

Answer 27

Mhm sec, I am getting - infinity for that

Answer 28

Oh oops, solve for x haha.

Answer 29

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

Answer 30

so x=0

Answer 31

Right, now for gy solve for y

Answer 32

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

Answer 33

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.

Answer 34

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

Answer 35

Get rid of it :P