JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Application of Derivatives

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 :

Answer: (B)

Solution

Surface area:

S = 2 (2 x \cdot 4 x + 4 x \cdot 5 x + 5 x \cdot 2 x) + 3 π r^{2} = 76 x^{2} + 3 π r^{2} = k

(constant).

Volume:

V = (2 x) (4 x) (5 x) + \frac{2}{3} π r^{3} = 40 x^{3} + \frac{2}{3} π r^{3}

.

From the constraint:

r^{2} = \frac{k - 76 x ^{2}}{3 π}

, so

r = (\frac{k - 76 x ^{2}}{3 π})^{1/2}

.

Substitute into

V

and differentiate with respect to

x

\frac{d V}{d x} = 120 x^{2} - \frac{152 x}{3} (\frac{k - 76 x ^{2}}{3 π})^{1/2}

Setting

\frac{d V}{d x} = 0

and simplifying:

120 x^{2} = \frac{152 x}{3} (\frac{k - 76 x ^{2}}{3 π})^{1/2}

\frac{45 x}{19} = (\frac{k - 76 x ^{2}}{3 π})^{1/2} = r

Squaring:

(\frac{45}{19})^{2} x^{2} = r^{2}

Therefore,

\frac{x}{r} = \frac{19}{45}

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Single correct option