Question

Use the second Derivative Test to find any local extrema of the function f(x,y) =e^(4x-x^(2)-y^(2))
Please explain step by step

Answer 1

\[f(x,y) = e ^{4x-x ^{2}-y ^{2}}\]

Answer 2

But, you know something about the thery? Because you have to calculate the differential matrix, then check what kind of diferrential matrix you got...and so on.

Answer 3

can you explain how to do the problem?

Answer 4

differential matrix? .... nah

Answer 5

what do you know about derivative rules?

Answer 6

i got fx = e^4x-x^2-y^2(4-2x)
and fy= 4x-x^2-y^2(2y)

Answer 7

but i dont know what to do after that

Answer 8

well, youve got good form. with the tiny little exponent i didnt see that ys :)

take Fxx and Fyy and Fxy i believe to find the extrema of the surface

Answer 9

I dont know how to get any of those in my book it says to set fx=0 and fy=0 and solve?

Answer 10

you know how to find Fx and Fy; find the x again of Fx to get Fxx, find the y again of Fy to get Fyy

Answer 11

i understand you get fxx and fyy by taking the derivative of fx and fy respectively but i keep getting the wrong answer

Answer 12

the wrong derivatives, or the wrong answer?

Answer 13

f(x,y) = x^2y^3 then fx = 2x.y^2

Then fxy = 2x.2y

Answer 14

both

Answer 15

let me try it; if anything just for the practice :)

Answer 16

\[f(x,y) = e ^{4x-x ^{2}-y ^{2}}\]

\[fx=(4-2x)e^{4x-x ^{2}-y ^{2}};\ fy=(-2y)e^{4x-x ^{2}-y ^{2}}\]

\[fxx=(-2)e^{4x-x ^{2}-y ^{2}}+(4-2x)^2e^{4x-x ^{2}-y ^{2}}\]\[
\ fyy=(-2)e^{4x-x ^{2}-y ^{2}}+(-2y)^2e^{4x-x ^{2}-y ^{2}}\]

good so far?

Answer 17

now, the zeros of this would give us some critical values, but with a 3d surface we might be stuck in a saddle point and so we have a way to check for that

Answer 18

Don´t forget that your critical values will be given by the gradient (fx, fy) = 0, 0

Answer 19

D=Fxx*Fyy - (Fxy)^2

Answer 20

i'm having trouble comming up with fxx and fyy did you distribute the 4-2x and 2y into e^4x^2x^2-2y?

Answer 21

no, i just used it as a product and pulled it thru the product rule

Answer 22

[fg]'=f'g+fg'

Answer 23

ohallright that makes total sense x.x

Answer 24

http://mysite.science.uottawa.ca/cstar050/1300notes/lecture19.pdf

this might be a good review if you need it

Answer 25

thank you, it seems really complicated though to multiply fxxfyy and subtract it from fxy^2

Answer 26

wolframalpha.com would help to ease that burden ;)

Answer 27

\[fx=(4-2x)e^{4x-x ^{2}-y ^{2}}\]
since 4-2x has no y in it; its considered a constant and we get:
\[fxy=(-2y)(4-2x)e^{4x-x ^{2}-y ^{2}}\]
\[(fxy)^2=((-2y)(4-2x)e^{4x-x ^{2}-y ^{2}})^2\]
\[(fxy)^2=(4y^2)(16-16x+4x^2)e^{(4x-x ^{2}-y ^{2})^2}\]
\[fxx*fyy=(14-16x+4x^2)(-2+4y^2)e^{(4x-x ^{2}-y ^{2})^2}\]


\[fxx*fyy=(14-16x+4x^2)(-2+4y^2)e^{(4x-x ^{2}-y ^{2})^2}\]\[-(fxy)^2=-(4y^2)(16-16x+4x^2)e^{(4x-x ^{2}-y ^{2})^2}\]
\[[(14-16x+4x^2)(-2+4y^2)-(4y^2)(16-16x+4x^2)]e^{(4x-x ^{2}-y ^{2})^2}\]

whew!! it just takes some time

Answer 28

its super long =/ but doesnt plugging in 0 for that give me 14(-2)-0?

Answer 29

if you plug in 0,0; yes

Answer 30

how does that compare to Fxx?

Answer 31

but that would mean that it is -28<0 so there is no max or min at (0,0) and the answer is e^4 =local max

Answer 32

and at fxx(0,0) i get 14which is >0

Answer 33

D > 0 and fxx(a,b) < 0 ; MAX
D > 0 and fxx(a,b) > 0 ; MIN
D<0; SADDLE

Answer 34

at (1,1) im getting -12e^4 for D and for fxx 2e^2 i dont think thats right

Answer 35

fx=0 when x=2
fy=0 when y=0 ; our critical point is at (2,0)

Answer 36

i get -2e^4 for fxx =/

Answer 37

\[fxx(2,0)=−2e^{4}\]\[fxy(2,0)=0\]

Answer 38

D = (14−16(2)+4(4))(−2)
D = (14−32+16)(−2)
D = (-2)(−2) = 4

D>0, and Fxx < 0 this would be a max right?

Answer 39

graphing it would be nice :)

Answer 40

the max is e^4 isnt that supposed to be our answer?

Answer 41

e^4 might be the max, but graphing it tells us that the max does indeed occur at (2,0)

e^(4x-x^2-y^2); (2,1) = e^4

Answer 42

(2,0) ... my keyboard hates me

Answer 43

lol dw its not you it's this problem it hates xD allright thanks so much! I know it was a headache, but I appreciate the help!

Answer 44

good luck with it all ;)