and and .

I find x=√(3/15) but the book says x=√0.6 How is that? See my calculations here: 2012-09-12 11.15.02.jpg

PS. This is after looking at ∂f/∂x=∂f/∂y=0 which leads to the point (0,0).

Results 1 to 5 of 5

- Sep 12th 2012, 01:21 AM #1

- Joined
- Sep 2012
- From
- sthlm
- Posts
- 53

## g'(x) = 0

and and .

I find x=√(3/15) but the book says x=√0.6 How is that? See my calculations here: 2012-09-12 11.15.02.jpg

PS. This is after looking at ∂f/∂x=∂f/∂y=0 which leads to the point (0,0).

- Sep 12th 2012, 02:47 AM #2

- Joined
- Jan 2008
- From
- UK
- Posts
- 484
- Thanks
- 66

- Sep 12th 2012, 05:34 AM #3

- Sep 12th 2012, 07:00 AM #4

- Joined
- Apr 2005
- Posts
- 19,027
- Thanks
- 2765

## Re: g'(x) = 0

Your title says "g'(x)= 0" but there is no "g" in your problem. In fact, there is no function of x only, no matter what you call it! Please tell us what the problem

**really**says. I suspect that you are looking for maximum and minimum values of f inside and on the circle of radius 3. You are right that the only point**inside**the circle, for which the partial derivatives are 0. ON the circle (and (0, 0) is not even on the circle), the simplest thing is to use a parameter: let , . Then . Differentiate that, with respect to and set that equal to 0.

- Sep 12th 2012, 07:01 AM #5

- Joined
- Sep 2012
- From
- sthlm
- Posts
- 53