If the radius of a cylinder was shrunk down to a quarter of its orignal size and the height was reduced to a third of its original size, what would be the formula to find the modified surface area?

The modified area is (1/48) (2πr (4h+3r))

Step-by-step explanation:

Let the radius be 'r' and height be 'h'.

Area of cylinder = 2π r (h+r)

The radius is shrunk down to quarter of its original radius

r = r/4

The height is reduced to a third of its original height

h = h/3

New Area = 2π (r/4) [ (h/3) + (r/4) ]

= (1/4) 2πr[ (4h+3r) / 12]

= (1/48) (2πr (4h+3r))