Question

Find all the local maxima, local minima and saddle points of the function.
f(x,y) = x e^y

Answer 1

@ganeshie8

Answer 2

fx(x,y)=e^y fy(x,y)=xe^y

e^y=0 x.e^y=0

y is undefined x=0

How do i continue?

Answer 3

fx(x,y) = e^y
fx(1,0) = e^0 =1
fy(x,y) = x e^y
fy(1,0) = (1)e^0 =1
f(1,0) = 1

Answer 4

f(x,y) = 1 + 1(x-1) + 1 (y-0)
= 1+(x-1)+y
= x+y
--------------------------------------...

Answer 5

Where did 1,0 come from?

Answer 6

Second-degree Taylor polynomial
Follow the example given in the link.

I have computed the derivative part.

Second degree Taylor polynomials pf f(x,y) at the point (1,0)

Answer 7

∂f/∂x = e^y
∂f/∂y = x e^y
∂^2f/∂x^2 = 0
∂^2f/∂y^2 = x e^y
∂^2f/∂x∂y = e^y

Answer 8

test for a saddle point is

fxx * fyy - (fxy)^2 < 0

Answer 9

I know that phi but as y is undefined how am i supposed to continue with this?

Answer 10

exactly .

Answer 11

fxx(x,y)=0,fxy(x,y)=xe^y,fyy(x,y)=e^y

Iv'e also calculated those things.

Answer 12

I'd suggest you go back and quickly review the CRITERIA to be used in determining whether or not you have a saddle point or a maximum or minimum.

Would you please defend y our statement, "y is undefined x=0" ?

Answer 13

e^y=0
Apply ln to both sides
ln(e^y) = ln(0)
y = undefined
since ln(0) is undefined

Answer 14

x.e^y=0
x=0

Answer 15

You don't need my help. You have a team of mathletes helping you here @hba ...

Answer 16

Let's re-visit the original f(x,y) = x e^y (or f(x,y) = x*e^y ):

\[f _{x}=e^y;~f _{xx}=0;~f _{y}=x*e^y;~ f _{yy}=x*e^y\]

Answer 17

Please look those partial derivatives over and ask yourself whether you agree or disagree.
I don't see that you need to apply the 'ln' operator here.

Answer 18

I agree with you 100 % @mathmale but we have to put fx=0 and fy=0 and get (x,y) you see

Answer 19

So at that point we can determine local maxima, local minima and saddle points using this:
fxx * fyy - (fxy)^2 < 0

Answer 20

Forget the <0 thing

Answer 21

you have a degenerate case, and it looks like there is no max,min or saddle. (though I don't know how to prove this)

Answer 22

Ok, let me see, if I remember correctly, if you have (a,b), which is a critical point of f(x,y), and the SECOND ORDER PARTIAL DERIV are continuous on the given region, then:

D=D(a,b)=f\(\sf_{xx}\)(a,b)f\(\sf_{yy}\)(a,b)-[f\(\sf_{xy}\)(a,b)]\(^2\)

If D>0 and f\(\sf_{xx}\) > 0, its relative min @ (a,b)
if D>0 and \(\sf_{xx}\) < 0, there theres a relative max @ (a,b)
if D<0 then the point (a,b) is a saddle point
D = -, then try something else

Answer 23

D=0 then test is conclusive

Answer 24

I'm sure you remember from Calc 1 how to find ur critical points.

Answer 25

Yes, sorry, D= 0*

Answer 26

@abb0t If y wasn't undefined i would have done it

Answer 27

@amistre64

Answer 28

Dis calculus doe