So the bottom radius is equal to √ (20 2-h 2)?

Volume v = π (400-h 2) * h/3?

Is to find the maximum value of (400-h 2) * h?

And 0?

f(h)=(400-h^2)*h=-h^3+400h？

f'(h)=-3h^2+400=0？

h= 20√3/3？

0? And then 0? 0, f(h) increases?

20√3/3? So x=20√3/3 is the maximum?

It is also the maximum value in the interval.