Consider a cuboid of sides 2 x , 4 x and 5 x and a closed hemisphere of radius r . If the sum of their surface areas is constant k , then the ratio x : r , for which the sum of their volumes is maximum, is

Step-by-step solution

Let sum of surface area k = 3 π r 2 + 76 x 2 ∵     k is constant so d k d x = 0 ⇒   6 π r d r d x + 2 × 76 x = 0 ⇒   d r d x = - 76 x 3 π r           ⋯ 1 Now total volume V = 2 3 π r 3 + 40 x 3 For maximum volume d V d x = 0 ⇒ 2 π r 2 d r d x + 120 x 2 = 0           ⋯ 2 ⇒   2 π r 2 - 76 x 3 π r + 120 x 2 = 0 ⇒   8 x - 19 r 3 + 15 x = 0 ⇒