Question

f(x,y)= (x^2 -y^2) e^((-x^2 -y^2)/2) Find absolute minima or maxima and saddle point

Answer 1

Find partial x and partial y

Answer 2

thnx but i amlookig for help

Answer 3

LOL, but for real, that is how you do it, for example this is what i got for partial x:

\[xe ^{(-x-y^2)/(2)}(-x^2+y^2+2)\]

Answer 4

The for partial y i got:

\[ye ^{(-x^2-y^2)/(2)}(-x^2+y^2-2)\]

Answer 5

Now we have to set both partial derivaitves to zero

Answer 6

f of x is 2x e^(-x^2 -y^2)/2 + (x^2 -y^2)e^(-x^2 -y^2 /2) (-x)

Answer 7

okay, that is what i have expecet that from what you got, we can factor out an:\[xe ^{(-x^2-y^2)/(2)}\]

Answer 8

and for partial y you will also be able to factor out:


\[ye ^{(-x^2-y^2)/(2)}\]

Answer 9

ok thanx

Answer 10

Then we should notice that, for both partial derivaitves, x and y=0 will make the derivaitve 0

Answer 11

but there are still yet more critical points to find

Answer 12

hm i get it.

Answer 13

no i dont think you do let me continue

Answer 14

i will solve therest thanx

Answer 15

so, what we can do is plug in x=0 into the partial y derivative (or vice versa, we could plug in the y=0 into the partial x dreivaitve), and we should end up with:

\[ye^ {-y^2/2}(y^2-2)\]

Answer 16

notice here that we get another zero, y=plus or minus sqrt(2)

Answer 17

We can then say that, y=plus or minus sqrt(2), will may partial y 0 when x=0. Thus plus or minus sqrt(2) makes partial x =0. So, (0, -sqrt(2)) and (0,sqrt(2) makes partial y and partial x 0

Answer 18

Now we can repeat this process using partial x, and we should end up with:
\[xe^{-x^2/2}(-x^2+2)\]

Answer 19

This will yield, two more critical points, (sqrt(2),0) and (-sqrt(2),0)