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
