Let say radius is r
its height is h
its lateral area = y
y = 2 pi r h
since the cylinder is inscribed in the sphere
So (2r )^2 + h^2 = 64
then 4 (r^2) = 64 - h^2
since y^2 = 4 (pi)^2 r^2 h^2
then y^2 = (pi)^2 *h^2 * (64 -h^2)
y^2 = 64 (pi)^2 * h^2 - (pi)^2 * h^4
2 y y' = 128 (pi)^2 * h - 4 (pi)^2 * h^3
putting y' = 0
4 (pi)^2 h ( 32 - h^2)=0
ether h = 0 testing this value (changing of the sign of y' before and after ) y is minimum
or h = 4 sqrt(2)
testing this value (changing of the sign of y' before and after ) y is maximum
So the maximum value of y^2 = (pi)^2 *32 *( 64 - 32)
y^2 = (pi)^2 * (32)^2
y = 32 (pi) square feet
hope this helps
