A.Maths

assume the height of cone = H
consider the cross-section through vertex
tan a = H/36

consider the cylinder
height h = (36-r)/tan a = (36-r) (36/H) = 36(36 - r)/H
curved surface area A = 2(pi)rh
=2(pi)r 36(36-r)/H
=72(pi)(36r- r^2)/H
for max. curved surface area,
dA/dr = 0
72(pi)(36-2r)/H = 0
36-2r = 0
r=18

d^2 A / dr^2 = 72(pi)(-2)/H < 0 which denotes the max. value for dA/dr = 0

Therefore, the curved surface area of the cylinder is maximum when the radius is 18cm.