JEE Advanced 2019 Paper 1 · Mathematics · Differentiation and Application of Derivatives · medium

A right circular cone of maximum volume is inscribed in a sphere of radius 3. A right circular cylinder of maximum volume is then inscribed in this…

Numerical answer

A right circular cone of maximum volume is inscribed in a sphere of radius

3

. A right circular cylinder of maximum volume is then inscribed in this cone, with its axis along the axis of the cone and one of its bases on the base of the cone. If

V

is the volume of the cylinder, find the value of

V/\pi

. Give the answer correct to two decimal places.

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Answer: 4.74

For a cone of height

h

in a sphere of radius

R

, the base radius satisfies

r^2=h(2R-h)

. Its volume

\frac\pi3h^2(2R-h)

is maximal when

h=\frac{4R}3

; for

R=3

this gives

h=4

and

r^2=8

. For a cylinder of radius

\rho

in this cone, its height is

4\left(1-\frac\rho r\right)

, so

V=\pi\rho^2\cdot4\left(1-\frac\rho r\right)

. Setting

\frac{dV}{d\rho}=0

gives

2\rho-\frac{3\rho^2}r=0

, i.e.

\rho=\frac{2r}3

. Then the cylinder height is

\frac43

and

V=\pi\cdot\frac{4r^2}{9}\cdot\frac43=\pi\cdot\frac{4\cdot8}{9}\cdot\frac43=\frac{128\pi}{27}

. Therefore

V/\pi=\frac{128}{27}\approx4.74

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