Determine the absolute maximum of y=x(r^2 + x^2)^ -2/3

[wind]

Hi, I am having trouble with this question, can someone pleas check over my work? Thanks.
Determine the absolute maximum of y=x(r^2 + x^2)^ -2/3 on the interval xe[0,r], where r is a constant.

y=x(r^2 + x^2)^ -2/3
y’= (r^2+x^2)^ -2/3 + (x)[-2/3(r^2+x^2)•2x]
y’= (r^2+x^2)^ -2/3 + (x)[-4/3(r^2+x^2)]
y’= (r^2+x^2)^ -2/3 + (x)[-4/3r^2- 4/3x^2)]
y’= (r^2+x^2)^ -2/3 + -4/3r^2x- 4/3x^2x
0= (r^2+x^2)^ -2/3 + -4/3r^2x- 4/3x^2x

How do I factor this?

Thanks

 

[tkhunny]

First: You're not very close on the derivative.
Second: Why do you need to factor anything?

After obtaining SOME correct form of the derivative, with a little algebra you can probably ignore the denominator.

\(\displaystyle y\;=\;x*(r^{2}\;+\;x^{2})^{(-2/3)}\)

\(\displaystyle y'\;=\;(r^{2}+x^{2})^{(-2/3)}\;+\;x*(-2/3)*(r^{2}+x^{2})^{(-5/3)}*(2x)\)

Do you see the part you missed? This leads to a nicely simplified:

\(\displaystyle y'\;=\;\frac{(3r^2\;-\;x^2)}{3*(r^2\;+\;x^2)^{(5/3)}}\)

Now what?

 

[wind]

set the numerator to o?

o=3r^2 - x^2
x^2=3r^2
x=root3r^2

like that?

 

[tkhunny]

1) I don't quite believe you worked through the details of that derivative. It will do you much good to get better at the algebra.

2) There is a zero (0) on the keyboard. It's right up next to the 9.

3) Yes, you can solve for 'x', but you have not yet assured yourself that it is, indeed, a proper solution. If that same value also makes the denominator zero, then you have a problem. (I realize I used the word "ignore", but I was partially kidding. ) If that value for x is outside the restricted Domain you started with, you have a different problem. Don't quit until you actually and deliberately answer the question that is asked. It is a common error to get some result and go home happy. Don't do it. Answer the real question.