Question

Find the relative extrema for the function.
f(x,y)=4x/x^2+y^2+1

Answer 1

@stamp

Answer 2

http://tutorial.math.lamar.edu/Classes/CalcIII/RelativeExtrema.aspx

Answer 3

already looked at that. I'm looking for a partial derivative check because its messy.

Answer 4

\[f(x,y)=\frac{4x}{x^2+y^2+1}\]

Answer 5

The second partial gets messy*

Answer 6

Does this look relevant?

http://www.wolframalpha.com/input/?i=solve+partial+derivative+of+4x%2F%28x%5E2%2By%5E2%2B1%29%3D0

Answer 7

ok so I solved partial with respect to x on paper and it checks with wolfram. Not sure how to find the critical point though. Do we need the second order partial?\[f_x=\frac{-4x^2+4y^2+4}{(x^2+y^2+1)^2}\]

Answer 8

yeah I think I got the 2nd partial finally because I re-wrote the equation so I could use product rule instead of quotient rule.

Answer 9

I am working on solving fxx

Answer 10

\[\frac{ (-4x^2+4y^2+1)(-4x) }{ (x^2+y^2+1)^3 }-\frac{ 8x }{ (x^2+y^2+1)^2}\]

Answer 11

I have been trying to work this out on paper and It has been destroying me, how did you rearrange it for the product rule?

Answer 12

yeah the same thing happened to me when I tried it the first time. I set \[f_x=(-4x^2+4y^2+4)*(x^2+y^2+1)^-2\]

Answer 13

my final answer also has an error but its very minor

Answer 14

\[f_{xx}=\frac{8x(x^2-3y^2-3)}{(x^2+y^2+1)^3}\]So what do we do with it

Answer 15

do we have to find\[f_y,\ f_{yy},\ f_{xy}\]

Answer 16

I will type out what i got

Answer 17

\[f_y=\frac{ -8xy }{ (x^2+y^2+1)^2 }\]

Answer 18

\[f_\left( yy \right)= \frac{ 32xy^2 }{ (x^2+y^2+1)^3 }-\frac{ 8y }{ (x^2+y^2+1)^2 }\]

Answer 19

\[f_ \left( yx \right)=\frac{ 32x^2y }{ (x^2+y^2+1)^3 }-\frac{ 8y }{ (x^2+y^2+1)^2 }\]

Answer 20

my yy should say -8x

Answer 21

I simplified my fxx and fyy to share a denominator, yours are separated.

but ok we have fx, fxx, fy, fyy.

What now

Answer 22

set a system to solve fx and fy = 0

Answer 23

This is the part where I get confused if you go to the lamar notes they explain how to find the relative min and maxes but I'm not sure if I'm doing it right

Answer 24

your first link now makes a bit of sense to me now

Answer 25

first partial x: http://www.wolframalpha.com/input/?i=partial+derivative+x+of+4x%2F%28x%5E2%2By%5E2%2B1%29

second partial x: http://www.wolframalpha.com/input/?i=partial+derivative+x+of+%284-4+x%5E2%2B4+y%5E2%29%2F%281%2Bx%5E2%2By%5E2%29%5E2

first partial y: http://www.wolframalpha.com/input/?i=partial+derivative+y+of+4x%2F%28x%5E2%2By%5E2%2B1%29

second partial y: http://www.wolframalpha.com/input/?i=partial+derivative+y+of+-8xy%2F%28x%5E2%2By%5E2%2B1%29%5E2

Answer 26

well all our partials , when simplified into one fraction, have the same denominators so we just set the numerators equal and solve for the numerators = 0

Answer 27

check my links to see the partials simplified

Answer 28

man i have been at work this whole time i will check this out later when i have free time