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.