JEE Main 2022 · 26 Jun · Shift 2 · Mathematics Q6

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MathematicsSingle correct+4 / -1Question 6 of 90

Cosider a cuboid of sides $2x$ , $4x$ and $5x$ 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 :

2:5

19:45

3:8

19:15

Answer and worked solution

Official answer: B

Constraint. Cuboid area $2(8+10+20)x^2=76x^2$ ; closed hemisphere $3\pi r^2$ : $76x^2+3\pi r^2=k$ .

Volume. $V=40x^3+\frac23\pi r^3$ .

Lagrange condition. $\dfrac{120x^2}{152x}=\dfrac{2\pi r^2}{6\pi r}$ , i.e. $\dfrac{15x}{19}=\dfrac r3$ , so $x:r=19:45$ .

(B)

Topics:Maxima and minima

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JEE Main 2022 · 26 Jun · Shift 2, Mathematics question 6