what's the maximum volume of a cone with a 10 cm generator i have no idea how to solve this i tried writing down the volume expression depending on the angle $ v= \frac{1}{3}\pi r^2h $ knowing that $ r= 10\sin\theta $ and $h= 10\cos\theta $ then $ V= \frac{1000}{3}\pi(\sin^2\theta \cos\theta) $ and that way the volume would deppend on the angle of the generator to the axis and i thought to differentiate the function of the volume in order to find the maximum value $ \frac{dv}{d\theta}= \frac{1000\pi}{3}\sin\theta ( 3 \cos^3\theta -1) $ this is simplifying at much as i can but i can't see a way to solve it for zero

The angle seems fine to me as a variable, but since it the formula contains $r^2$ the function should be $v(\theta)=\cfrac {1000\pi}3\sin^2 \theta \cos \theta$. Then you need to take care over the differentiation.
Now $\sin^2 \theta \cos \theta=(1-\cos^2\theta) \cos \theta=\cos \theta -\cos^3\theta$
So $\cfrac {dv}{d \theta}=\cfrac {1000\pi}3\left(-\sin \theta +3\sin\theta\cos^2\theta\right)$
And this expression is zero if $\sin \theta (3\cos^2\theta-1)=0$, which happens if $\sin \theta =0$ or $\cos \theta =\pm\cfrac 1{\sqrt 3}$. You just have to identify the maximum.
