• August 14th 2010, 09:05 PM
$v= {10 \pi r^2(6-r) /3$

the whole is meant to be over 3
• August 14th 2010, 09:06 PM
Prove It
You have

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

Start by expanding the brackets. You should be able to find the derivative term by term.
• August 14th 2010, 09:16 PM
is it
$v' = \frac{ 60 \pi r^2 - 10 \pi r^3 }{3} =0$

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

$r= 6 , 0$ zero is excluded as 0 for a length (in the original question cannot make sense)
• August 14th 2010, 09:19 PM
Prove It

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

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

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

$v = 20\pi r^2 - \frac{10\pi}{3} r^3$
$v' = 40 \pi r - 10 \pi r^2$ for stat points v' = 0
$10 \pi r (4 - r) = 0$
therefore
$r= 0 or 4$

0 doesnt make sense so 4 only?
• August 14th 2010, 09:35 PM
Prove It
But it should be possible to say that if you don't have a radius you don't have a volume, so $r = 0$ is definitely a minimum...