Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere

Solution:

Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R.

 

 

Let V be the volume of the cone.

Then, V = 1/3πr²h

Height of the cone is given by,

h = R + AB

= R + √R² - r²  [ABC is a right triangle]

Hence,

V = 1/3π r²(R + √R² - r²)

= 1/3πr² + 1/3π r² (√R² - r²)

dV/dr = 2/3πrR + 2/3π r √R² - r² - 1/3π r² - (- 2r)/√R² - r²

= 2/3πrR + (2πr (R² - r²) - πr³)/3√R² - r²

= 2/3πrR + (2πrR - 3πr³)/3√R² - r²

d²V/dr² = 2/3πR + [3√R² - r² (2π R² - 9π r²) - (2π rR² - 3πr³). (- 2r)/(6√R² - r²)] / 9(R² - r²)

= 2/3πR + [9(R² - r²) (2πrR² - 9πr²) + 2πr²R² + 3πr⁴] / 27(R² - r²)^3/2

Now,

dV/dr = 0

⇒ 2/3πrR + [2πrR² - 3πr³] / [3√R² - r²] = 0

⇒ 2/3π rR = [3πr³ - 2πrR²] / [3√R² - r²]

⇒ 2R = (3r² - 2R²) / (√R² - r²)

⇒ 2R√R² - r² = 3r² - 2R²

⇒ 4R² (R² - r ²) = (3r² - 2R²)²

⇒ 4R⁴ - 4R²r² = 9r⁴ + 4R⁴ - 12r ²R²

⇒ 9r⁴ = 8R²r²

⇒ r² = 8/9 R²

When, r² = 8/9 R²

Then,

dV/dr² < 0

By second derivative test, the volume of the cone is the maximum when r² = 8/9 R²

When, r² = 8/9 R²

Then,

h = R + √R² - 8/9 R²

= R + √1/9 R²

= R + R/3

= 4/3 R

Therefore,

V = 1/3π (8/9 R²) (4/3 R)

= 4/27 (4/3π R³)

= 8/27 (volume of sphere)

Hence, the volume of the largest cone that can be inscribed in the sphere is 8/27 the volume of the sphere

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 23

Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere.

Summary:

Hence we have proved that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere