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

Well, if you let the independent variable be $h$, then you have $r^2+h^2=100$, so that $\frac13\pi r^2h$ becomes $\frac13\pi(100-h^2)h$, and $dV/dh=\frac13\pi(100-3h^2)$, set that to zero and you’re on your way.
