# Homework Help: Find and Classify the critical points of f(x,y)

1. May 20, 2012

### knowLittle

1. The problem statement, all variables and given/known data
f(x,y)= 16(y^2) +(x^4) y + 4(x^2) + 4

My problem is recognizing which critical points to consider valuables.

2. Relevant equations
fxx, fyy, fxy, and second partials test.
D=fxx(fyy)- (fxy)^2

3. The attempt at a solution
I found:
fx=0
4(x^3)y +8x=0
(x^2) y= -2
Now, from here I can discern that either
x=+-1 , y= -2 **this solution is not considered in the solutions manual. Anyone care to explain**
OR
x=2, y= -(1/2) OR
x= -2, y= -1/2
Also, I was thinking about plugging in values of the other fy=0 part. It gets even more complicated.

Now:
fy=0
32y + (x^4) =0
y= -(x^4)/32
From here the only thing I thought about was
x=0, y=0.
I also thought about solving for y and replacing on the other fx equation, but things don't look good and the solutions shows that it's wrong.

Can anyone give me tips to discern critical points?

Thank you.

2. May 20, 2012

### knowLittle

Nevermind. Thank you. I found a solution.

The values of x and y have to fit in both fx and fy. Also, replacing y (- (x^4)/32) in fx will yield the solutions:
x=2, y=(-1/2)
x= -2, y= -1/2

3. May 20, 2012

### knowLittle

A question still remains, why is x=0, y=0 considered? It does not fit with both equations.
It only fits in fy.

I found an explanation, but I might be wrong. I am often.
It comes from computing the first partial:
fx=0
4(x^3)y+8x=0
x(4(x^2)y+ 8) =0, here I can say that x=0 and plug this value of x in fy.

Here fy=0:
32y+x^4=0
32y=-x^4
y= -(x^4)/32, if x=0--> y=0. So, x=0 and y=0 is also a critical point.

Is this correct?

Last edited: May 20, 2012
4. May 20, 2012

### knowLittle

Delete this post. Sorry.

Last edited: May 20, 2012
5. May 20, 2012

### sharks

You can use "Edit:" on a new line, followed by whatever you want to add.

6. May 20, 2012

### Ray Vickson

As a matter of presentation (and to prevent yourself from making errors), _first_ write down both f_x and f_y, _then_ deal with the equations f_x = 0 and f_y = 0. The problem with first writing f_x = 0 and manipulating it is that you might accidentally drop one or more of the possible roots (as you did above). If you are a bit more systematic it will help you in the long run.

RGV

7. May 20, 2012

### knowLittle

I don't know which roots I missed.
f_x= 4(x^3)y + 8x
f_y=32 y + x^4

8. May 20, 2012

### Ray Vickson

f_x = 0 --> 4*x*(y*x^2 + 2) = 0, so either x = 0 or y*x^2 + 2 = 0. When you did it originally you missed the root x = 0. You found it again, later---but that is the point: it is easy to go astray if you are not systematic. Being systematic also makes *marking* your work easier, which means that the marker is less likely to take off points because they cannot follow what you are doing.

RGV

9. May 20, 2012

### knowLittle

Ok. Thank you.

So, my reasoning is correct for the final result. Right?