$\displaystyle v= {10 \pi r^2(6-r) /3 $

the whole is meant to be over 3

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- Aug 14th 2010, 09:05 PMlink0099quickly differentiate this please
$\displaystyle v= {10 \pi r^2(6-r) /3 $

the whole is meant to be over 3 - Aug 14th 2010, 09:06 PMProve It
You have

$\displaystyle v = \frac{10\pi r^2(6-r)}{3}$.

Start by expanding the brackets. You should be able to find the derivative term by term. - Aug 14th 2010, 09:16 PMlink0099
is it

$\displaystyle v' = \frac{ 60 \pi r^2 - 10 \pi r^3 }{3} =0$

$\displaystyle v' = \pi r^2(20 - \frac{ 10}{3}r) =0 $

$\displaystyle r= 6 , 0 $ zero is excluded as 0 for a length (in the original question cannot make sense) - Aug 14th 2010, 09:19 PMProve It
No, your expansion gives

$\displaystyle v = \frac{60\pi r^2 - 10\pi r^3}{3}$

$\displaystyle = 20\pi r^2 - \frac{10\pi}{3} r^3$

but you haven't differentiated it yet... - Aug 14th 2010, 09:26 PMlink0099
ok i really gotta practice with pi in these eqn's.

$\displaystyle v = 20\pi r^2 - \frac{10\pi}{3} r^3 $

$\displaystyle v' = 40 \pi r - 10 \pi r^2 $ for stat points v' = 0

$\displaystyle 10 \pi r (4 - r) = 0 $

therefore

$\displaystyle r= 0 or 4 $

0 doesnt make sense so 4 only? - Aug 14th 2010, 09:35 PMProve It
Your calculations are correct.

But it should be possible to say that if you don't have a radius you don't have a volume, so $\displaystyle r = 0$ is definitely a minimum...