Question

f(x,y)=xye^(-x-y) find local min, max or if it is a saddle point

Answer 1

fx=(x-1)y(-e^(-x-y))
fy=x(y-1)(-e^(-x-y))

Answer 2

i have a trouble finding x and y values

Answer 3

it is only 0,1?

Answer 4

yep
x=y=0 and x=y=1

Answer 5

now just find D=(fxx)(fyy)-(fxy)^2 for those points

Answer 6

thats the problem though :p

Answer 7

why, can't find fxx, fyy, and/or fxy ??

Answer 8

so i got fxx=e^(-x-y), fyy=e^(-x-y)+y-1 and fxy=e^(-x-y)+y-1

Answer 9

D(x,y)=(e^(-x-y)+y)(e^(-x-y)+y-1) -(e^(-x-y)+y-1)^2

Answer 10

then i get D(a,b) for (0,0) = 0 so its inconclusive? then for (1,1)= 1 but for fxx(a,b)=2 so its a local min?

Answer 11

(1,1) is not D=1 I don't think

Answer 12

does the derivative look right?

Answer 13

\[\text{Sorry for the delay, I'm in Calculus I.}\\
\text{We're given }f(x,y)=xy e^{-x-y}.\text{ The stationary points are those such that}\\
\ \ \ \nabla f(x^*,y^*)=0\\
\text{Let us determine the partial derivatives of the function.}\\
\ \ \ f_x=-y(e^{-x-y})(x-1) \\
\ \ \ f_y=-x(e^{-x-y})(y-1)\\
\ \ \ f_{xx}=y(e^{-x-y})(x-2)\\
\ \ \ f_{yy}=x(e^{-x-y})(y-2)\\
\ \ \ f_{xy}=(x-1)(y-1)(e^{-x-y})\\
\text{Let's find the stationary points first, and then classify them.}\\
\ \ \ f_x=f_y=0\\
\ \ \ -y(e^{-x-y})(x-1)=-x(e^{-x-y})(y-1)=0\\
\ \ \ -y(x-1)=-x(y-1)=0\\
\ \ \ -xy + y = -xy + x = 0\\
\ \ \ y = x = xy\\
\text{So our two stationary points are (0,0) and (1,1). Now, let's attempt to classify them.}\\
\ \ \ D=f_{xx}f_{yy}-[f_{xy}]^2\\
\ \ \ \ \ \ =xye^{-2x-2y}(x-2)(y-2)-(x-1)^2(y-1)^2e^{-2x-2y}\\
\ \ \ \ \ \ =(e^{-2x-2y})[xy(x-2)(y-2)-(x-1)^2(y-1)^2]\\
\ \ \ D(0,0)=(e^0)[0-0]=0\\
\text{therefore our test is inconclusive for (0,0).}\\
\ \ \ D(1,1)=(e^{-4})[1-0]=e^{-4}>0\\
\ \ \ f_{xx}(1,1)=1(e^{-2})(-1)=-e^{-2}<0\\
\text{therefore (1,1) is a relative maximum.}
\]

Answer 14

thanks a lot,

Answer 15

i messed up the derivatives

Answer 16

I'm not entirely sure how to determine what (0,0) is though... the second derivative test isn't enough to classify it. WolframAlpha says it's a saddle-point but I don't know why.