g'(x) = 0

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September 12th 2012, 01:21 AM
jacob93

g'(x) = 0

$f(x,y)=4x^3-xy^2$ and $x^2+y^2 \leq 9$ and $x \geq 0$ .

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 AM
a tutor

Re: g'(x) = 0

What was the question?
September 12th 2012, 05:34 AM
kalyanram

Re: g'(x) = 0

You made a substitution mistake. Substituted $y^2 = 3-x^2$ instead of $y^2 = 9-x^2$ . See below for solution.

Spoiler:

$g(x) = x(4x^2 - (9-x^2)) = 5x^3 - 9x \implies g'(x) = 15x^2 - 9 = 0$ $\implies x= \pm\sqrt{\frac{3}{5}}$ as we have $x \ge 0, so \hspace{3mm} x = \sqrt{\frac{3}{5}} = \sqrt{0.6}$
September 12th 2012, 07:00 AM
HallsofIvy

Re: g'(x) = 0

Quote:

Originally Posted by jacob93

$f(x,y)=4x^3-xy^2$ and $x^2+y^2 \leq 9$ and $x \geq 0$ .

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).

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 $x= 3 cos(\theta)$ , $y= 3 sin(\theta)$ . Then $f(x, y)= f(3cos(\theta), 3sin(\theta))= 4r^3cos^3(\theta)- r^3cos(\theta)sin^2(\theta)$ . Differentiate that, with respect to $\theta$ and set that equal to 0.
September 12th 2012, 07:01 AM
jacob93

Re: g'(x) = 0

Thanks