Find Critical Points of f(x,y)

[Shmoo]

Homework Statement

Determine the critical points f(x,y) = x^3y + xy

Homework Equations

The Attempt at a Solution

fx(x,y) = 3x^2 + y
3x^2y+y=0
y(3x^2 +1) = 0
y = 0
3x^2=1
x=1/3fy(x,y) = x^3 + x
x^3 + x = 0
x(x^2+1) = 0
x=0
x^2= -1
x= -1

I don't know how to take it from here as I have only seen this done where you substitute y values in equations for x. What are the critical points of the above? Thanks for the help!

 

[lanedance]

so all viable combinations of the solutions you found above, where fx = fy = 0, are critical points

if you want to exmaine the nature of the critical point you will need to consider a type of 2nd derivative test or the hessian matrix

 

[HallsofIvy]

Homework Statement

Determine the critical points f(x,y) = x^3y + xy

Homework Equations

The Attempt at a Solution

fx(x,y) = 3x^2 + y
3x^2y+y=0
y(3x^2 +1) = 0
y = 0
3x^2=1
x=1/3

So either y= 0 or x= 1/3.

fy(x,y) = x^3 + x
x^3 + x = 0
x(x^2+1) = 0
x=0
x^2= -1
x= -1

In particular, x= 1/3 does not satisfy this. In order that both equations be satisfied, you must have x= 0, y= 0 or x= -1, y= 0.

I don't know how to take it from here as I have only seen this done where you substitute y values in equations for x. What are the critical points of the above? Thanks for the help!

 

[ehild]

The Attempt at a Solution

y(3x^2 +1) = 0
y = 0
3x^2=1
x=1/3

I suppose that you assumed 3x^2+1=0. From here, everything is wrong what you did. From 3x^2+1=0 does not follow that 3x^2=1.
From 3x^2=1 does not come that x=1/3.

fy(x,y) = x^3 + x
x^3 + x = 0
x(x^2+1) = 0
x=0
x^2= -1
x= -1

X^2=-1 does not mean that x=-1. Do you know what x^2 means? Can be the square of any real number equal to a negative number?


ehild