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).

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- September 12th 2012, 01:21 AMjacob93g'(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). - September 12th 2012, 02:47 AMa tutorRe: g'(x) = 0
What was the question?

- September 12th 2012, 05:34 AMkalyanramRe: g'(x) = 0
You made a substitution mistake. Substituted instead of . See below for solution.

__Spoiler__: - September 12th 2012, 07:00 AMHallsofIvyRe: 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. - September 12th 2012, 07:01 AMjacob93Re: g'(x) = 0
Thanks