Question

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 a constant k , then the ratio x : r , for which the sum of their volumes is maximum, is

• A. 2 : 5
• B. 19 : 45
• C. 3 : 8
• D. 19 : 15

Answer 1

Answer: (B)

19 : 45

Answer 2

The correct answer is (B) : 19 : 45
∵ s1 + s2 = k
76x² + 3πr² = k
∴ 152x $\frac{dx}{dr }$+ 6πr = 0
∴ $\frac{dx}{dr }$_ _ = $\frac{-6πr}{152x}$
Now
V = $40x^3$+ $\frac{2}{3}$πr³
∴ $\frac{dx}{dr }$ = 120x² . $\frac{dx}{dr }$ + 2πr² = 0
⇒ 120x² . ( $\frac{-6π}{152}$ $\frac{r}{x}$ ) + 2πr² = 0
⇒ 120 $(\frac{x}{r})$ = $2π$ $(\frac{152}{6π})$
⇒ $(\frac{x}{r})$ = $\frac{152}{3}$ $\frac{1}{120}$ = $\frac{19}{45}$