Find all relative maxima/minima and saddle points

Homework Statement

f(x,y)=2x^3+xy^2+5x^2+y^2

Homework Equations

The Attempt at a Solution

fx(x,y)=6x^2+y^2+25x
fy(x,y)=2xy+2y
fxx(x,y)=12x+25
fyy(x,y)=2y+2
fxy(x,y)=2y
I found the partial derivatives from the equation. I am stuck at finding the critical points from the first two fx(x,y) and fy(x,y). I equaled 0=6x^2+y^2+25x and solved for y^2. I then got y^2=-6x^2-25x and tried substituting it in, fy(x,y)= 2x(6-x^2-25x)^2+2(-6x^2-25x)^2. which did not work out well as I was able to do in similar problems.

 

[LCKurtz]

Homework Statement

f(x,y)=2x^3+xy^2+5x^2+y^2

Homework Equations

The Attempt at a Solution

fx(x,y)=6x^2+y^2+25x
fy(x,y)=2xy+2y
fxx(x,y)=12x+25
fyy(x,y)=2y+2
fxy(x,y)=2y
I found the partial derivatives from the equation. I am stuck at finding the critical points from the first two fx(x,y) and fy(x,y). I equaled 0=6x^2+y^2+25x and solved for y^2. I then got y^2=-6x^2-25x and tried substituting it in, fy(x,y)= 2x(6-x^2-25x)^2+2(-6x^2-25x)^2. which did not work out well as I was able to do in similar problems.

You have $f_y = 2xy+2y$. Set that equal to $0$, and factor it getting $2y(x+1)=0$. That tells you there are two possibilities, either $y=0$ or $x=-1$. Go from there.
Edit: Added, check your arithmetic for $f_x$ too.

 

You have $f_y = 2xy+2y$. Set that equal to $0$, and factor it getting $2y(x+1)=0$. That tells you there are two possibilities, either $y=0$ or $x=-1$. Go from there.
Edit: Added, check your arithmetic for $f_x$ too.

that was a mistype on the fx either than that thank you.