# Find all relative maxima/minima and saddle points

## Homework Statement

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

## 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.

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## Homework Statement

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

## 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.

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.